Section 3.3 Convergence Tests
It is very common to encounter series for which it is difficult, or even virtually impossible, to determine the sum exactly. Often you try to evaluate the sum approximately by truncating it, i.e. having the index run only up to some finite rather than infinity. But there is no point in doing so if the series diverges. So you like to at least know if the series converges or diverges. Furthermore you would also like to know what error is introduced when you approximate by the “truncated series” That’s called the truncation error. There are a number of “convergence tests” to help you with this.
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The authors should be a little more careful making such a blanket statement. While it is true that it is not wise to approximate a divergent series by taking terms with large, there are cases when one can get a very good approximation by taking terms with small! For example, the Taylor remainder theorem shows us that when the derivative of a function grows very quickly with Taylor polynomials of degree with large, can give bad approximations of while the Taylor polynomials of degree one or two can still provide very good approximations of when is very small. As an example of this, one of the triumphs of quantum electrodynamics, namely the computation of the anomalous magnetic moment of the electron, depends on precisely this. A number of important quantities were predicted using the first few terms of divergent power series. When those quantities were measured experimentally, the predictions turned out to be incredibly accurate.
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The field of asymptotic analysis often makes use of the first few terms of divergent series to generate approximate solutions to problems; this, along with numerical computations, is one of the most important techniques in applied mathematics. Indeed, there is a whole wonderful book (which, unfortunately, is too advanced for most Calculus 2 students) devoted to playing with divergent series called, unsurprisingly, “Divergent Series” by G.H. Hardy. This is not to be confused with the “Divergent” series by V. Roth set in a post-apocalyptic dystopian Chicago. That latter series diverges quite dramatically from mathematical topics, while the former does not have a film adaptation (yet).
Subsection 3.3.1 The Divergence Test
Our first test is very easy to apply, but it is also rarely useful. It just allows us to quickly reject some “trivially divergent” series. It is based on the observation that
- by definition, a series
converges to when the partial sums converge to - Then, as
we have and, because too, we also have - So
This tells us that, if we already know that a given series is convergent, then the term of the series, must converge to as tends to infinity. In this form, the test is not so useful. However the contrapositive of the statement is a useful test for divergence.
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We have discussed the contrapositive a few times in the CLP notes, but it doesn’t hurt to discuss it again here (or for the reader to quickly look up the relevant footnote in Section 1.3 of the CLP-1 text). At any rate, given a statement of the form “If A is true, then B is true” the contrapositive is “If B is not true, then A is not true”. The two statements in quotation marks are logically equivalent — if one is true, then so is the other. In the present context we have “If ( converges) then ( converges to ).” The contrapositive of this statement is then “If ( does not converge to 0) then ( does not converge).”
Example 3.3.2. A simple divergence.
Let Then
So the series diverges.
Warning 3.3.3.
The divergence test is a “one way test”. It tells us that if is nonzero, or fails to exist, then the series diverges. But it tells us absolutely nothing when In particular, it is perfectly possible for a series to diverge even though An example is We’ll show in Example 3.3.6, below, that it diverges.
Now while convergence or divergence of series like can be determined using some clever tricks — see the optional §3.3.9 —, it would be much better to have methods that are more systematic and rely less on being sneaky. Over the next subsections we will discuss several methods for testing series for convergence.
Note that while these tests will tell us whether or not a series converges, they do not (except in rare cases) tell us what the series adds up to. For example, the test we will see in the next subsection tells us quite immediately that the series
converges. However it does not tell us its value .
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This series converges to Apéry’s constant The constant is named for Roger Apéry (1916–1994) who proved that this number must be irrational. This number appears in many contexts including the following cute fact — the reciprocal of Apéry’s constant gives the probability that three positive integers, chosen at random, do not share a common prime factor.
Subsection 3.3.2 The Integral Test
In the integral test, we think of a series that we cannot evaluate explicitly, as the area of a union of rectangles, with representing the area of a rectangle of width one and height Then we compare that area with the area represented by an integral, that we can evaluate explicitly, much as we did in Theorem 1.12.17, the comparison test for improper integrals. We’ll start with a simple example, to illustrate the idea. Then we’ll move on to a formulation of the test in general.
Example 3.3.4. Convergence of the harmonic series.
Visualise the terms of the harmonic series as a bar graph — each term is a rectangle of height and width The limit of the series is then the limiting area of this union of rectangles. Consider the sketch on the left below.
It shows that the area of the shaded columns, is bigger than the area under the curve with That is
If we were to continue drawing the columns all the way out to infinity, then we would have
We are able to compute this improper integral exactly:
That is the area under the curve diverges to and so the area represented by the columns must also diverge to
It should be clear that the above argument can be quite easily generalised. For example the same argument holds mutatis mutandis for the series
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Latin for “Once the necessary changes are made”. This phrase still gets used a little, but these days mathematicians tend to write something equivalent in English. Indeed, English is pretty much the lingua franca for mathematical publishing. Quidquid erit.
Indeed we see from the sketch on the right above that
and hence
This last improper integral is easy to evaluate:
Thus we know that
and so the series must converge.
The above arguments are formalised in the following theorem.
Theorem 3.3.5. The Integral Test.
Let be any natural number. If is a function which is defined and continuous for all and which obeys
for all and decreases as increases and for all
Proof.
Let be any fixed integer with Then
converges if and only if converges — removing a fixed finite number of terms from a series cannot impact whether or not it converges.- Since
for all the sequence of partial sums obeys That is, increases as increases. - So
must either converge to some finite number or increase to infinity. That is, either converges to a finite number or it is
Look at the figure above. The shaded area in the figure is because
- the first shaded rectangle has height
and width and hence area and - the second shaded rectangle has height
and width and hence area and so on
This shaded area is smaller than the area under the curve for So
and, if the integral is finite, the sum is finite too. Furthermore, the desired bound on the truncation error is just the special case of this inequality with
For the “divergence case” look at the figure above. The (new) shaded area in the figure is again because
- the first shaded rectangle has height
and width and hence area and - the second shaded rectangle has height
and width and hence area and so on
This time the shaded area is larger than the area under the curve for So
and, if the integral is infinite, the sum is infinite too.
Now that we have the integral test, it is straightforward to determine for which values of the series
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This series, viewed as a function of is called the Riemann zeta function, or the Euler-Riemann zeta function. It is extremely important because of its connections to prime numbers (among many other things). Indeed Euler proved that Riemann showed the connections between the zeros of this function (over complex numbers ) and the distribution of prime numbers. Arguably the most famous unsolved problem in mathematics, the Riemann hypothesis, concerns the locations of zeros of this function.
converges.
Example 3.3.6. The test: .
Let We’ll now use the integral test to determine whether or not the series (which is sometimes called the -series) converges.
- To do so, we need a function
that obeys for all bigger than some Certainly obeys for all So let’s pick this and try (We can always increase later if we need to.) - This function also obeys the other two conditions of Theorem 3.3.5:
for all and decreases as increases because for all
- So the integral test tells us that the series
converges if and only if the integral converges.
So we conclude that converges if and only if This is sometimes called the -test.
- In particular, the series
which is called the harmonic series, has and so diverges. As we add more and more terms of this series together, the terms we add, namely get smaller and smaller and tend to zero, but they tend to zero so slowly that the full sum is still infinite. - On the other hand, the series
has and so converges. This time as we add more and more terms of this series together, the terms we add, namely tend to zero (just) fast enough that the full sum is finite. Mind you, for this example, the convergence takes place very slowly — you have to take a huge number of terms to get a decent approximation to the full sum. If we approximate by the truncated series we make an error of at most but really slowly.
We now know that the dividing line between convergence and divergence of occurs at We can dig a little deeper and ask ourselves how much more quickly than the term needs to shrink in order for the series to converge. We know that for large the function is smaller than for any positive — you can convince yourself of this with a quick application of L’Hôpital’s rule. So it is not unreasonable to ask whether the series
converges. Notice that we sum from because when And we don’t need to stop there . We can analyse the convergence of this sum with any power of
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We could go even further and see what happens if we include powers of and other more exotic slow growing functions.
Example 3.3.7. .
Let We’ll now use the integral test to determine whether or not the series converges.
- As in the last example, we start by choosing a function that obeys
for all bigger than some Certainly obeys for all So let’s use that and try - Now let’s check the other two conditions of Theorem 3.3.5:
- Both
and are positive for all so for all - As
increases both and increase and so increases and decreases.
- So the integral test tells us that the series
converges if and only if the integral converges. - To test the convergence of the integral, we make the substitution
and hence the integral converges if and only if
So we conclude that converges if and only if
Subsection 3.3.3 The Comparison Test
Our next convergence test is the comparison test. It is much like the comparison test for improper integrals (see Theorem 1.12.17) and is true for much the same reasons. The rough idea is quite simple. A sum of larger terms must be bigger than a sum of smaller terms. So if we know the big sum converges, then the small sum must converge too. On the other hand, if we know the small sum diverges, then the big sum must also diverge. Formalising this idea gives the following theorem.
Theorem 3.3.8. The Comparison Test.
“Proof”.
We will not prove this theorem here. We’ll just observe that it is very reasonable. That’s why there are quotation marks around “Proof”. For an actual proof see the optional section 3.3.10.
- If
converges to a finite number and if the terms in are smaller than the terms in then it is no surprise that converges too. - If
diverges (i.e. adds up to ) and if the terms in are larger than the terms in then of course adds up to and so diverges, too.
The comparison test for series is also used in much the same way as is the comparison test for improper integrals. Of course, one needs a good series to compare against, and often the series (from Example 3.3.6), for some turns out to be just what is needed.
Example 3.3.9. .
We could determine whether or not the series converges by applying the integral test. But it is not worth the effort . Whether or not any series converges is determined by the behaviour of the summand for very large So the first step in tackling such a problem is to develop some intuition about the behaviour of when is very large.
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Go back and quickly scan Theorem 3.3.5; to apply it we need to show that is positive and decreasing (it is), and then we need to integrate To do that we reread the notes on partial fractions, then rewrite and so and then arctangent appears, etc etc. Urgh. Okay — let’s go back to the text now and see how to avoid this.
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To understand this consider any series We can always cut such a series into two parts — pick some huge number like Then The first sum, though it could be humongous, is finite. So the left hand side, is a well-defined finite number if and only if is a well-defined finite number. The convergence or divergence of the series is determined by the second sum, which only contains for “large”
- Step 1: Develop intuition. In this case, when
is very large10
The symbol “ ” means “much larger than”. Similarly, the symbol “ ” means “much less than”. Good shorthand symbols can be quite expressive. so that We already know, from Example 3.3.6, that converges if and only if So which has converges, and we would expect that converges too. - Step 2: Verify intuition. We can use the comparison test to confirm that this is indeed the case. For any
so that So the comparison test, Theorem 3.3.8, with and tells us that converges.
Of course the previous example was “rigged” to give an easy application of the comparison test. It is often relatively easy, using arguments like those in Example 3.3.9, to find a “simple” series with almost the same as when is large. However it is pretty rare that for all It is much more common that for some constant This is enough to allow application of the comparison test. Here is an example.
Example 3.3.10. .
As in the previous example, the first step is to develop some intuition about the behaviour of when is very large.
- Step 1: Develop intuition. When
is very large, so that the numerator and so that the denominator
So when is very large - Step 2: Verify intuition. We can use the comparison test to confirm that this is indeed the case. To do so we need to find a constant
such that is smaller than for all A good wayto do that is to factor the dominant term (in this case11
This is very similar to how we computed limits at infinity way way back near the beginning of CLP-1. ) out of the numerator and also factor the dominant term (in this case ) out of the denominator.So now we need to find a constant such that is smaller than for all- First consider the numerator
For all and
So the numerator is always smaller than - Next consider the denominator
- When
lies between and so that is between and and consequently is between and
- As the numerator
is always smaller than and is always smaller than the fraction
We now know thatand, since we know converges, the comparison test tells us that converges.
The last example was actually a relatively simple application of the comparison theorem — finding a suitable constant can be really tedious . Fortunately, there is a variant of the comparison test that completely eliminates the need to explicitly find
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Really, really tedious. And you thought some of those partial fractions computations were bad …
The idea behind this isn’t too complicated. We have already seen that the convergence or divergence of a series depends not on its first few terms, but just on what happens when is really large. Consequently, if we can work out how the series terms behave for really big then we can work out if the series converges. So instead of comparing the terms of our series for all just compare them when is big.
Theorem 3.3.11. Limit Comparison Theorem.
Proof.
(a) Because we are told that we know that,
- when
is large, is very close to so that is very close to - In particular, there is some natural number
so that for all and hence with for all- The comparison Theorem 3.3.8 now implies that
converges.
(b) Let’s suppose that (If just replace with ) Because we are told that we know that,
- when
is large, is very close to - In particular, there is some natural number
so that and hence with for all- The comparison Theorem 3.3.8 now implies that
diverges.
The next two examples illustrate how much of an improvement the above theorem is over the straight comparison test (though of course, we needed the comparison test to develop the limit comparison test).
Example 3.3.12. .
Set We first try to develop some intuition about the behaviour of for large and then we confirm that our intuition was correct.
- Step 1: Develop intuition. When
the numerator and the denominator so that and it looks like our series should converge by Example 3.3.6 with - Step 2: Verify intuition. To confirm our intuition we set
and compute the limit converges by Example 3.3.6 with So our series converges by the limit comparison test, Theorem 3.3.11.
Example 3.3.13. again.
We can also try to deal with the series of Example 3.3.12, using the comparison test directly. But that requires us to find so that
We might do this by examining the numerator and denominator separately:
- The numerator isn’t too bad since for all
- The denominator is quite a bit more tricky, since we need a lower bound, rather than an upper bound, and we cannot just write
which is false. Instead we have to make a more careful argument. In particular, we’d like to find and so that i.e. for all For we have So for
Putting the numerator and denominator back together we have
and the comparison test then tells us that our series converges. It is pretty clear that the approach of Example 3.3.12 was much more straightforward.
Subsection 3.3.4 The Alternating Series Test
When the signs of successive terms in a series alternate between and like for example in the series is called an alternating series. More generally, the series
is alternating if every Often (but not always) the terms in alternating series get successively smaller. That is, then In this case:
- The first partial sum is
- The second partial sum,
is smaller than by - The third partial sum,
is bigger than by but because remains smaller than See the figure below. - The fourth partial sum,
is smaller than by but because remains bigger than Again, see the figure below. - And so on.
So the successive partial sums oscillate, but with ever decreasing amplitude. If, in addition, tends to as tends to the amplitude of oscillation tends to zero and the sequence converges to some limit
This is illustrated in the figure
Here is a convergence test for alternating series that exploits this structure, and that is really easy to apply.
Theorem 3.3.14. Alternating Series Test.
“Proof”.
We shall only give part of the proof here. For the rest of the proof see the optional section 3.3.10. We shall fix any natural number and concentrate on the last statement, which gives a bound on the truncation error (which is the error introduced when you approximate the full series by the partial sum )
This is of course another series. We’re going to study the partial sums
for that series.
- If
with even, for all both even and odd. - Similarly, if
with odd, for all for all both even and odd.
So we now know that lies between its first term, and for all While we are not going to prove it here (see the optional section 3.3.10), this implies that, since as the series converges and that
lies between and
Example 3.3.15. Convergence of the alternating harmonic series.
We have already seen, in Example 3.3.6, that the harmonic series diverges. On the other hand, the series converges by the alternating series test with Note that
for all so that really is an alternating series, and decreases as increases, and
so that all of the hypotheses of the alternating series test, i.e. of Theorem 3.3.14, are satisfied. We shall see, in Example 3.5.20, that
Example 3.3.16. .
You may already know that In any event, we shall prove this in Example 3.6.5, below. In particular
is an alternating series and satisfies all of the conditions of the alternating series test, Theorem 3.3.14a:
- The terms in the series alternate in sign.
- The magnitude of the
term in the series decreases monotonically as increases. - The
term in the series converges to zero as
So the alternating series test guarantees that, if we approximate, for example,
then the error in this approximation lies between and the next term in the series, which is That is
so that
which, to seven decimal places says
(To seven decimal places )
The alternating series test tells us that, for any natural number the error that we make when we approximate by the partial sum has magnitude no larger than This tends to zero spectacularly quickly as increases, simply because increases spectacularly quickly as increases . For example
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The interested reader may wish to check out “Stirling’s approximation”, which says that
Example 3.3.17. Computing .
We will shortly see, in Example 3.5.20, that if then
Suppose that we have to compute to within an accuracy of Since we can get by evaluating at so that
By the alternating series test, this series converges. Also by the alternating series test, approximating by throwing away all but the first terms
introduces an error whose magnitude is no more than the magnitude of the first term that we threw away.
To achieve an error that is no more than we have to choose so that
The best way to do so is simply to guess — we are not going to be able to manipulate the inequality into the form and even if we could, it would not be worth the effort. We need to choose so that the denominator is at least That is easy, because the denominator contains the factor which is at least whenever i.e. whenever So we will achieve an error of less than if we choose
This is not the smallest possible choice of but in practice that just doesn’t matter — your computer is not going to care whether or not you ask it to compute a few extra terms. If you really need the smallest that obeys you can next just try then and so on.
So in this problem, the smallest acceptable
Subsection 3.3.5 The Ratio Test
The idea behind the ratio test comes from a reexamination of the geometric series. Recall that the geometric series
converges when and diverges otherwise. So the convergence of this series is completely determined by the number This number is just the ratio of successive terms — that is
In general the ratio of successive terms of a series, is not constant, but depends on However, as we have noted above, the convergence of a series is determined by the behaviour of its terms when is large. In this way, the behaviour of this ratio when is small tells us nothing about the convergence of the series, but the limit of the ratio as does. This is the basis of the ratio test.
Theorem 3.3.18. Ratio Test.
Warning 3.3.19.
Proof.
(a) Pick any number obeying We are assuming that approaches as In particular there must be some natural number so that for all So for all In particular
for all The series is a geometric series with ratio smaller than one in magnitude and so converges. Consequently, by the comparison test with replaced by and replaced by the series converges. So the series converges too.
(b) We are assuming that approaches as In particular there must be some natural number so that for all So for all That is, increases as increases as long as So for all and cannot converge to zero as So the series diverges by the divergence test.
Example 3.3.20. .
Fix any two nonzero real numbers and We have already seen in Example 3.2.4 and Lemma 3.2.5 — we have just renamed to — that the geometric series converges when and diverges when We are now going to consider a new series, constructed by differentiating each term in the geometric series This new series is
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We shall see later, in Theorem 3.5.13, that the function is indeed the derivative of the function Of course, such a statement only makes sense where these series converge — how can you differentiate a divergent series? (This is not an allusion to a popular series of dystopian novels.) Actually, there is quite a bit of interesting and useful mathematics involving divergent series, but it is well beyond the scope of this course.
Let’s apply the ratio test.
The ratio test now tells us that the series converges if and diverges if It says nothing about the cases But in both of those cases does not converge to zero as and the series diverges by the divergence test.
Notice that in the above example, we had to apply another convergence test in addition to the ratio test. This will be commonplace when we reach power series and Taylor series — the ratio test will tell us something like
Of course, we will still have to to determine what happens when To determine convergence or divergence in those cases we will need to use one of the other tests we have seen.
Example 3.3.21. .
Once again, fix any two nonzero real numbers and We again start with the geometric series but this time we construct a new series by integrating each term, from to giving The resulting new series is
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We shall also see later, in Theorem 3.5.13, that the function is indeed an antiderivative of the function
To apply the ratio test we need to compute
The ratio test now tells us that the series converges if and diverges if It says nothing about the cases
If the series reduces to
which is just times the harmonic series, which we know diverges, by Example 3.3.6.
In conclusion, the series converges if and only if
The ratio test is often quite easy to apply, but one must always be careful when the limit of the ratio is The next example illustrates this.
Example 3.3.22. .
In this example, we are going to see three different series that all have One is going to diverge and the other two are going to converge.
- The first series is the harmonic series
- The second series is the alternating harmonic series
- The third series is
that this series converges. But it also has
Let’s do a somewhat artificial example that forces us to combine a few of the techniques we have seen.
Example 3.3.23. .
Again, the convergence of this series will depend on
- Let us start with the ratio test — so we compute
So in the limit as
we are left with - The ratio test then tells us that if
the series diverges, while when the series converges. - This leaves us with the cases
and - Setting
gives the series decreases as increases. So we define the function in with ) and verify that is a decreasing function of To prove that, it suffices to show its derivative is negative when this is negative and so is a decreasing function. Thus we can apply the alternating series test to show that the series converges when - When
the series becomes is large, the summand is approximately which suggests that the series will diverge by comparison with To formalise this, we can use the limit comparison theorem: diverges, we know that our series also diverges.
So in summary the series converges when and diverges otherwise.
Subsection 3.3.6 Convergence Test List
We now have half a dozen convergence tests:
- Divergence Test
- works well when the
term in the series fails to converge to zero as tends to infinity
- Alternating Series Test
- works well when successive terms in the series alternate in sign
- don’t forget to check that successive terms decrease in magnitude and tend to zero as
tends to infinity
- Integral Test
- works well when, if you substitute
for in the term you get a function, that you can integrate - don’t forget to check that
and that decreases as increases
- Ratio Test
- works well when
simplifies enough that you can easily compute - this often happens when
contains powers, like or factorials, like - don’t forget that
tells you nothing about the convergence/divergence of the series
- Comparison Test and Limit Comparison Test
- works well when, for very large
the term is approximately the same as a simpler term (see Example 3.3.10) and it is easy to determine whether or not converges - don’t forget to check that
- usually the Limit Comparison Test is easier to apply than the Comparison Test
Subsection 3.3.7 Optional — The Leaning Tower of Books
Imagine that you are about to stack a bunch of identical books on a table. But you don’t want to just stack them exactly vertically. You want to built a “leaning tower of books” that overhangs the edge of the table as much as possible.
How big an overhang can you get? The answer to that question, which we’ll now derive, uses a series!
-
Let’s start by just putting book #1 on the table. It’s the red book labelled “
” in the figure below.Use a horizontal -axis with corresponding to the right hand edge of the table. Imagine that we have placed book #1 so that its right hand edge overhangs the end of the table by a distance- In order for the book to not topple off of the table, we need its centre of mass to lie above the table. That is, we need the
-coordinate of the centre mass of which we shall denote to obey will be exactly half way between the right hand end of the book, which is at and the left hand end of the book, which is at So
Thus book #1 does not topple off of the table provided -
Now let’s put books #1 and #2 on the table, with the right hand edge of book #1 at
and the right hand edge of book #2 at as in the figure below.- In order for book #2 to not topple off of book #1, we need the centre of mass of book #2 to lie above book #1. That is, we need the
-coordinate of the centre mass of which is to obey -
Assuming that book #2 does not topple off of book #1, we still need to arrange that the pair of books does not topple off of the table. Think of the pair of books as the combined red object in the figureIn order for the combined red object to not topple off of the table, we need the centre of mass of the combined red object to lie above the table. That is, we need the
-coordinate of the centre mass of the combined red object, which we shall denote to obeyThe centre of mass of the combined red object is the weighted averageof the centres of mass of16
It might be a good idea to review the beginning of §2.3 at this point. and As and have the same weight,and the combined red object does not topple off of the table if
In conclusion, our two-book tower survives ifIn particular we may choose and to satisfy and Then, substituting into gives -
Before considering the general “
-book tower”, let’s now put books #1, #2 and #3 on the table, with the right hand edge of book #1 at the right hand edge of book #2 at and the right hand edge of book #3 at as in the figure below.- In order for book #3 to not topple off of book #2, we need the centre of mass of book #3 to lie above book #2. That is, we need the
-coordinate of the centre mass of which is to obey -
Assuming that book #3 does not topple off of book #2, we still need to arrange that the pair of books, book #2 plus book #3 (the red object in the figure below), does not topple off of book #1.In order for this combined red object to not topple off of book #1, we need the
-coordinate of its centre mass, which we denote to obeyThe centre of mass of the combined red object is the weighted average of the centre of masses of and As and have the same weight,and the combined red object does not topple off of book #1 if -
Assuming that book #3 does not topple off of book #2, and also that the combined book #2 plus book #3 does not topple off of book #1, we still need to arrange that the whole tower of books, book #1 plus book #2 plus book #3 (the red object in the figure below), does not topple off of the table.In order for this combined red object to not topple off of the table, we need the
-coordinate of its centre mass, which we denote to obeyThe centre of mass of the combined red object is the weighted average of the centre of masses of and and As they all have the same weight,and the combined red object does not topple off of the table if
In conclusion, our three-book tower survives ifIn particular, we may choose and to satisfySubstituting the second equation into the first givesNext substituting the third equation into the second, and then using the formula above for givesand finally - We are finally ready for the general “
-book tower”. Stack books on the table, with book on the bottom and book at the top, and with the right hand edge of book # at The same centre of mass considerations as above show that the tower survives if ’s to obey from the second equation into the first equation gives from the third equation into the second equation gives This is times the partial sum of the harmonic series As we saw in Example 3.3.6 (the test), the harmonic series diverges. So, as goes to infinity also goes to infinity. We may make the overhang as largeas we like!17
At least if our table is strong enough.
Subsection 3.3.8 Optional — The Root Test
There is another test that is very similar in spirit to the ratio test. It also comes from a reexamination of the geometric series
The ratio test was based on the observation that which largely determines whether or not the series converges, could be found by computing the ratio The root test is based on the observation that can also be determined by looking that the root of the term with very large:
Of course, in general, the term is not exactly However, if for very large the term is approximately proportional to with given by the above limit, we would expect the series to converge when and diverge when That is indeed the case.
Theorem 3.3.24. Root Test.
Warning 3.3.25.
Beware that the root test provides absolutely no conclusion about the convergence or divergence of the series if
Proof.
(a) Pick any number obeying We are assuming that approaches as In particular there must be some natural number so that for all So for all and the series converges by comparison to the geometric series
(b) We are assuming that approaches (or grows unboundedly) as In particular there must be some natural number so that for all So for all and the series diverges by the divergence test.
Example 3.3.26. .
We have already used the ratio test, in Example 3.3.23, to show that this series converges when and diverges when We’ll now use the root test to draw the same conclusions.
- Write
- We compute
- We’ll now show that the limit of
as is exactly To do, so we first compute the limit of the logarithm. - So
and the root test also tells us that if the series diverges, while when the series converges.
We have done the last example once, in Example 3.3.23, using the ratio test and once, in Example 3.3.26, using the root test. It was clearly much easier to use the ratio test. Here is an example that is most easily handled by the root test.
Example 3.3.27. .
Write Then
Now we take the limit,
by Example 3.7.20 in the CLP-1 text with As the limit is strictly smaller than the series converges.
To draw the same conclusion using the ratio test, one would have to show that the limit of
as is strictly smaller than 1. It’s clearly better to stick with the root test.
Subsection 3.3.9 Optional — Harmonic and Basel Series
Subsubsection 3.3.9.1 The Harmonic Series
The series
that appeared in Warning 3.3.3, is called the Harmonic series , and its partial sums
18
The interested reader should use their favourite search engine to read more on the link between this series and musical harmonics. You can also find interesting links between the Harmonic series and the so-called “jeep problem” and also the problem of stacking a tower of dominoes to create an overhang that does not topple over.
are called the Harmonic numbers. Though these numbers have been studied at least as far back as Pythagoras, the divergence of the series was first proved in around 1350 by Nicholas Oresme (1320-5 – 1382), though the proof was lost for many years and rediscovered by Mengoli (1626–1686) and the Bernoulli brothers (Johann 1667–1748 and Jacob 1655–1705).
Oresme’s proof is beautiful and all the more remarkable that it was produced more than 300 years before calculus was developed by Newton and Leibnitz. It starts by grouping the terms of the harmonic series carefully:
So one can see that this is and so must diverge .
19
The grouping argument can be generalised further and the interested reader should look up Cauchy’s condensation test.
There are many variations on Oresme’s proof — for example, using groups of two or three. A rather different proof relies on the inequality
which follows immediately from the Taylor series for given in Theorem 3.6.7. From this we can bound the exponential of the Harmonic numbers:
Subsubsection 3.3.9.2 The Basel Problem
The problem of determining the exact value of the sum of the series
is called the Basel problem. The problem is named after the home town of Leonhard Euler, who solved it. One can use telescoping series to show that this series must converge. Notice that
Hence we can bound the partial sum:
Thus, as increases, the partial sum increases (the series is a sum of positive terms), but is always smaller than So the sequence of partial sums converges.
Mengoli posed the problem of evaluating the series exactly in 1644 and it was solved — not entirely rigorously — by Euler in 1734. A rigorous proof had to wait another 7 years. Euler used some extremely cunning observations and manipulations of the sine function to show that
He used the Maclaurin series
and a product formula for sine
Extracting the coefficient of from both expansions gives the desired result. The proof of the product formula is well beyond the scope of this course. But notice that at least the values of which make the left hand side of (✶) zero, namely with integer, are exactly the same as the values of which make the right hand side of (✶) zero .
20
Knowing that the left and right hand sides of (✶) are zero for the same values of is far from the end of the story. Two functions and having the same zeros, need not be equal. It is certainly possible that where is a function that is nowhere zero. The interested reader should look up the Weierstrass factorisation theorem.
This approach can also be used to compute for and show that they are rational multiples of The corresponding series of odd powers are significantly nastier and getting closed form expressions for them remains a famous open problem.
21
Search-engine your way to “Riemann zeta function”.
Subsection 3.3.10 Optional — Some Proofs
In this optional section we provide proofs of two convergence tests. We shall repeatedly use the fact that any sequence of real numbers which is increasing (i.e. for all ) and bounded (i.e. there is a constant such that for all ) converges. We shall not prove this fact .
22
It is one way to state a property of the real number system called “completeness”. The interested reader should use their favourite search engine to look up “completeness of the real numbers”.
We start with the comparison test, and then move on to the alternating series test.
Theorem 3.3.28. The Comparison Test (stated again).
Proof.
(a) By hypothesis converges. So it suffices to prove that converges, because then, by our Arithmetic of series Theorem 3.2.9,
will converge too. But for all so that, for all the partial sums
increase with but never gets bigger than the finite number So the partial sums converge as
(b) For all the partial sum
By hypothesis, and hence grows without bound as So as
Theorem 3.3.29. Alternating Series Test (stated again).
Proof.
Let be an even natural number. Then the partial sum obeys
and
So the sequence of even partial sums is a bounded, increasing sequence and hence converges to some real number Since and converges zero as the odd partial sums also converge to That is between and (the first dropped term) was already proved in §3.3.4.
Exercises 3.3.11 Exercises
Exercises — Stage 1 .
1.
Select the series below that diverge by the divergence test.
(A)
(B)
(C)
(D)
2.
Select the series below whose terms satisfy the conditions to apply the integral test.
(A)
(B)
(C)
(D)
3.
Suppose there is some threshold after which a person is considered old, and before which they are young.
4.
Below are graphs of two sequences with positive terms. Assume the sequences continue as shown. Fill in the table with conclusions that can be made from the direct comparison test, if any.
if |
if |
|
and if |
then |
then |
and if |
then |
then |
5.
For each pair of series below, decide whether the second series is a valid comparison series to determine the convergence of the first series, using the direct comparison test and/or the limit comparison test.
compared to the divergent series compared to the convergent series compared to the convergent series compared to the divergent series
6.
Suppose is a sequence with Does converge or diverge, or is it not possible to determine this from the information given? Why?
7.
8.
9.
What flaw renders the following reasoning invalid?
Q: Determine whether the seriesconverges or diverges. A: We want to compare this series to the seriesNote both this series and the series in the question have positive terms.
We note thatfor sufficiently large. Therefore, Therefore, Sinceand are both expressions that work out to be positive for the values of under consideration, we can divide both sides of the inequality by these terms without having to flip the inequality. So, Now, we claimconverges. Now, by the Direct Comparison Test, we conclude thatconverges.
10.
Which of the series below are alternating?
(A)
(B)
(C)
(D)
11.
Give an example of a convergent series for which the ratio test is inconclusive.
12.
Imagine you’re taking an exam, and you momentarily forget exactly how the inequality in the ratio test works. You remember there’s a ratio, but you don’t remember which term goes on top; you remember there’s something about the limit being greater than or less than one, but you don’t remember which way implies convergence.
should mean that the sum diverges (rather than converging).
13.
14. (✳).
Write down a sequence such that exists and is nonzero. (You don’t have to carry out the Limit Comparison Test)
15. (✳).
Decide whether each of the following statements is true or false. If false, provide a counterexample. If true provide a brief justification.
- If
then converges. - If
then converges. - If
and diverges, then diverges.
Exercises — Stage 2 .
16. (✳).
Does the series converge?
17. (✳).
Determine, with explanation, whether the series converges or diverges.
18. (✳).
Determine whether the series is convergent or divergent. If it is convergent, find its value.
19.
Does the following series converge or diverge?
20.
Evaluate the following series, or show that it diverges:
21.
Evaluate the following series, or show that it diverges:
22.
Does the following series converge or diverge?
23.
Does the following series converge or diverge?
24.
Does the following series converge or diverge?
25.
Evaluate the following series, or show that it diverges:
26.
Does the following series converge or diverge?
27.
Does the following series converge or diverge?
28. (✳).
Show that the series converges.
29. (✳).
30. (✳).
Does converge or diverge?
31. (✳).
32. (✳).
Determine whether the series converges.
33. (✳).
Does converge or diverge?
34. (✳).
Determine, with explanation, whether each of the following series converge or diverge.
35. (✳).
36. (✳).
Determine whether each of the following series converge or diverge.
37.
Evaluate the following series, or show that it diverges:
38. (✳).
Determine whether the series is convergent or divergent. If it is convergent, find its value.
39. (✳).
Determine, with explanation, whether each of the following series converge or diverge.
40. (✳).
Determine, with explanation, whether each of the following series converges or diverges.
41. (✳).
Determine whether the series is convergent or divergent.
42. (✳).
43. (✳).
It is known that (you don’t have to show this). Find so that the partial sum of the series, satisfies Be sure to say why your method can be applied to this particular series.
44. (✳).
The series converges to some number (you don’t have to prove this). According to the Alternating Series Estimation Theorem, what is the smallest value of for which the partial sum of the series is at most away from For this value of write out the partial sum of the series.
Exercises — Stage 3 .
45. (✳).
Determine, with explanation, whether the following series converge or diverge.
46. (✳).
(a) Prove that diverges.
(b) Explain why you cannot conclude that diverges from part (a) and the Integral Test.
(c) Determine, with explanation, whether converges or diverges.
47. (✳).
48. (✳).
49. (✳).
50. (✳).
51. (✳).
52.
Suppose the frequency of word use in a language has the following pattern: times, while the second-most-frequently used word should appear about times, and so on.
So, in a text of 100 words, we expect the most frequently used word to appear
If books written in this language use distinct words, then the most commonly used word accounts for roughly what percentage of total words used?
53.
Suppose the sizes of cities in a country adhere to the following pattern: if the largest city has population then the -th largest city has population
If the largest city in this country has 2 million people and the smallest city has 1 person, then the population of the entire country is (For many ’s in this sum is not an integer. Ignore that.) Evaluate this sum approximately, with an error of no more than 1 million people.