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CLP-2 Integral Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 3.2 Series

Subsection 3.2.1 Series

A series is a sum
\begin{gather*} a_1+a_2+a_3+\cdots+a_n+\cdots \end{gather*}
of infinitely many terms. In summation notation, it is written
\begin{gather*} \sum_{n=1}^\infty a_n \end{gather*}
You already have a lot of experience with series, though you might not realise it. When you write a number using its decimal expansion you are really expressing it as a series. Perhaps the simplest example of this is the decimal expansion of \(\frac{1}{3}\text{:}\)
\begin{align*} \frac{1}{3} &= 0.3333\cdots \end{align*}
Recall that the expansion written in this way actually means
\begin{align*} 0.333333\cdots &= \frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\frac{3}{10000}+\cdots =\sum_{n=1}^\infty\frac{3}{10^n} \end{align*}
The summation index \(n\) is of course a dummy index. You can use any symbol you like (within reason) for the summation index.
\begin{equation*} \sum_{n=1}^\infty\frac{3}{10^n} =\sum_{i=1}^\infty\frac{3}{10^i} =\sum_{j=1}^\infty\frac{3}{10^j} =\sum_{\ell=1}^\infty\frac{3}{10^\ell} \end{equation*}
A series can be expressed using summation notation in many different ways. For example the following expressions all represent the same series:
\begin{align*} \sum_{n=1}^\infty\frac{3}{10^n} &= \overbrace{\frac{3}{10}}^{n=1} +\overbrace{\frac{3}{100}}^{n=2} +\overbrace{\frac{3}{1000}}^{n=3}+\cdots\\ \sum_{j=2}^\infty\frac{3}{10^{j-1}} &= \overbrace{\frac{3}{10}}^{j=2} +\overbrace{\frac{3}{100}}^{j=3} +\overbrace{\frac{3}{1000}}^{j=4}+\cdots\\ \sum_{\ell=0}^\infty\frac{3}{10^{\ell+1}} &= \overbrace{\frac{3}{10}}^{\ell=0} +\overbrace{\frac{3}{100}}^{\ell=1} +\overbrace{\frac{3}{1000}}^{\ell=3}+\cdots\\ \frac{3}{10}+\sum_{n=2}^\infty\frac{3}{10^n} &= \frac{3}{10} +\overbrace{\frac{3}{100}}^{n=2} +\overbrace{\frac{3}{1000}}^{n=3}+\cdots \end{align*}
We can get from the first line to the second line by substituting \(n=j-1\) — don’t forget to also change the limits of summation (so that \(n=1\) becomes \(j-1=1\) which is rewritten as \(j=2\)). To get from the first line to the third line, substitute \(n=\ell+1\) everywhere, including in the limits of summation (so that \(n=1\) becomes \(\ell+1=1\) which is rewritten as \(\ell=0\)).
Whenever you are in doubt as to what series a summation notation expression represents, it is a good habit to write out the first few terms, just as we did above.
Of course, at this point, it is not clear whether the sum of infinitely many terms adds up to a finite number or not. In order to make sense of this we will recast the problem in terms of the convergence of sequences (hence the discussion of the previous section). Before we proceed more formally let us illustrate the basic idea with a few simple examples.

Example 3.2.1. \(\ds \sum_{n=1}^\infty\frac{3}{10^n}\).

As we have just seen above the series \(\sum_{n=1}^\infty\frac{3}{10^n}\) is
\begin{equation*} \sum_{n=1}^\infty\frac{3}{10^n} = \overbrace{\frac{3}{10}}^{n=1} +\overbrace{\frac{3}{100}}^{n=2} +\overbrace{\frac{3}{1000}}^{n=3}+\cdots \end{equation*}
Notice that the \(n^{\mathrm th}\) term in that sum is
\begin{equation*} 3\times 10^{-n} = 0.\!\!\overbrace{00\cdots 0}^{n-1\ \mathrm{zeroes}}\!3 \end{equation*}
So the sum of the first \(5\text{,}\) \(10\text{,}\) \(15\) and \(20\) terms in that series are
\begin{align*} \sum_{n=1}^5\frac{3}{10^n} &= 0.33333 & \sum_{n=1}^{10}\frac{3}{10^n} &= 0.3333333333\\ \sum_{n=1}^{15}\frac{3}{10^n} &= 0.333333333333333 & \sum_{n=1}^{20}\frac{3}{10^n} &= 0.33333333333333333333 \end{align*}
It sure looks like that, as we add more and more terms, we get closer and closer to \(0.\dot 3=\frac{1}{3}\text{.}\) So it is very reasonable
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Of course we are free to define the series to be whatever we want. The hard part is defining it to be something that makes sense and doesn’t lead to contradictions. We’ll get to a more systematic definition shortly.
to define \(\sum_{n=1}^\infty\frac{3}{10^n}\) to be \(\frac{1}{3}\text{.}\)

Example 3.2.2. \(\ds \sum_{n=1}^\infty 1\) and \(\sum_{n=1}^\infty (-1)^n\).

Every term in the series \(\sum_{n=1}^\infty 1\) is exactly \(1\text{.}\) So the sum of the first \(N\) terms is exactly \(N\text{.}\) As we add more and more terms this grows unboundedly. So it is very reasonable to say that the series \(\sum_{n=1}^\infty 1\) diverges.
The series
\begin{equation*} \sum_{n=1}^\infty (-1)^n =\overbrace{(-1)}^{n=1} +\overbrace{1}^{n=2} +\overbrace{(-1)}^{n=3} +\overbrace{1}^{n=4} +\overbrace{(-1)}^{n=5}+\cdots \end{equation*}
So the sum of the first \(N\) terms is \(0\) if \(N\) is even and \(-1\) if \(N\) is odd. As we add more and more terms from the series, the sum alternates between \(0\) and \(-1\) for ever and ever. So the sum of all infinitely many terms does not make any sense and it is again reasonable to say that the series \(\sum_{n=1}^\infty(-1)^n\) diverges.
In the above examples we have tried to understand the series by examining the sum of the first few terms and then extrapolating as we add in more and more terms. That is, we tried to sneak up on the infinite sum by looking at the limit of (partial) sums of the first few terms. This approach can be made into a more formal rigorous definition. More precisely, to define what is meant by the infinite sum \(\sum_{n=1}^\infty a_n\text{,}\) we approximate it by the sum of its first \(N\) terms and then take the limit as \(N\) tends to infinity.

Definition 3.2.3.

The \(N^{\rm th}\) partial sum of the series \(\sum_{n=1}^\infty a_n\) is the sum of its first \(N\) terms
\begin{gather*} S_N=\sum_{n=1}^N a_n. \end{gather*}
Conversely, for each \(N\ge 2\text{,}\) \(a_N=S_N-S_{N-1}\text{.}\)
The partial sums form a sequence \(\big\{S_N\big\}_{N=1}^\infty\text{.}\) If this sequence of partial sums converges \(S_N \to S\) as \(N\rightarrow\infty\) then we say that the series \(\sum_{n=1}^\infty a_n\) converges to \(S\) and we write
\begin{equation*} \sum_{n=1}^\infty a_n=S \end{equation*}
If the sequence of partial sums diverges, we say that the series diverges.
In the last definition, we introduced the notation \(S_N\) for the \(N^{\rm th}\) partial sum of a series \(\sum_{n=1}^\infty a_n\) that has the smallest value of the index \(n\) being \(1\text{.}\) It is sometimes convenient, as we shall see in the next example, to express a series with the index having a smallest value other than one, like for example \(\sum_{n=0}^\infty a_n\text{.}\) Then there are a couple of different commonly used definitions for the partial sum notation \(S_N\text{.}\) One is \(S_N=\sum_{n=0}^N a_n\text{,}\) which is the sum of all terms having indices less than or equal to \(N\text{.}\) With this definition, we still have \(a_N=S_N-S_{N-1}\text{.}\) A second commonly used definition is the sum of the first \(N\) terms in the series, i.e. \(S_N=\sum_{n=0}^{N-1} a_n\text{.}\) Under this definition \(a_N=S_{N+1}-S_N\text{.}\) You may of course use whatever definition you like. But you should state clearly what definition you are using.

Example 3.2.4. Geometric Series.

Let \(a\) and \(r\) be any two fixed real numbers with \(a\ne 0\text{.}\) The series
\begin{equation*} a + ar + ar^2 + \cdots + ar^n + \cdots = \sum_{n=0}^\infty ar^n \end{equation*}
is called the geometric series with first term \(a\) and ratio \(r\text{.}\)
Notice that we have chosen to start the summation index at \(n=0\text{.}\) That’s fine. The first
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It is actually quite common in computer science to think of \(0\) as the first integer. In that context, the set of natural numbers is defined to contain \(0\text{:}\) \(\mathbb{N} = \left\{0,1,2,\cdots \right\}\) while the notation \(\mathbb{Z}^+ = \left\{1,2,3,\cdots \right\}\) is used to denote the (strictly) positive integers. Remember that in this text, as is more standard in mathematics, we define the set of natural numbers to be the set of (strictly) positive integers.
term is the \(n=0\) term, which is \(ar^0=a\text{.}\) The second term is the \(n=1\) term, which is \(ar^1=ar\text{.}\) And so on. We could have also written the series \(\sum_{n=1}^\infty ar^{n-1}\text{.}\) That’s exactly the same series — the first term is \(ar^{n-1}\big|_{n=1}=ar^{1-1}=a\text{,}\) the second term is \(ar^{n-1}\big|_{n=2}=ar^{2-1}=ar\text{,}\) and so on
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This reminds the authors of the paradox of Hilbert’s hotel. The hotel with an infinite number of rooms is completely full, but can always accommodate one more guest. The interested reader should use their favourite search engine to find more information on this.
. Regardless of how we write the geometric series, \(a\) is the first term and \(r\) is the ratio between successive terms.
Geometric series have the extremely useful property that there is a very simple formula for their partial sums. Denote the partial sum by
\begin{align*} S_N &= \sum_{n=0}^Nar^n = a + ar + ar^2 + \cdots + ar^N.\\ \end{align*}

The secret to evaluating this sum is to see what happens when we multiply it by \(r\text{:}\)

\begin{align*} rS_N &= r\big(a + ar + ar^2 + \cdots + ar^N\big)\\ & = ar + ar^2 + ar^3 + \cdots + ar^{N+1} \end{align*}
Notice that this is almost the same
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One can find similar properties of other special series, that allow us, with some work, to cancel many terms in the partial sums. We will shortly see a good example of this. The interested reader should look up “creative telescoping” to see how this idea might be used more generally, though it is somewhat beyond this course.
as \(S_N\text{.}\) The only differences are that the first term, \(a\text{,}\) is missing and one additional term, \(ar^{N+1}\text{,}\) has been tacked on the end. So
\begin{align*} S_N &= a + ar + ar^2 + \cdots + ar^N\\ r S_N &= \phantom{ar +} ar + ar^2 + \cdots + ar^N + ar^{N+1}\\ \end{align*}

Hence taking the difference of these expressions cancels almost all the terms:

\begin{align*} (1-r) S_N &= a - ar ^{N+1} = a (1-r^{N+1})\\ \end{align*}

Provided \(r\neq 1\) we can divide both side by \(1-r\) to isolate \(S_N\text{:}\)

\begin{align*} S_N &= a \cdot \frac{1-r^{N+1}}{1-r}. \end{align*}
On the other hand, if \(r=1\text{,}\) then
\begin{align*} S_N &= \underbrace{a + a +\cdots + a}_{N+1 \text{ terms}} = a (N+1) \end{align*}
So in summary:
\begin{equation*} S_N = \begin{cases} a\frac{1-r^{N+1}}{1-r} & \text{if $r\ne 1$} \\ a(N+1) & \text{if $r=1$} \end{cases} \end{equation*}
Now that we have this expression we can determine whether or not the series converges. If \(|r| \lt 1\text{,}\) then \(r^{N+1}\) tends to zero as \(N\rightarrow\infty\text{,}\) so that \(S_N\) converges to \(\frac{a}{1-r}\) as \(N\rightarrow\infty\) and
\begin{equation*} \sum_{n=0}^\infty ar^n = \frac{a}{1-r} \text{provided $|r| \lt 1$}. \end{equation*}
On the other hand if \(|r|\ge 1\text{,}\) \(S_N\) diverges. To understand this divergence, consider the following 4 cases:
  • If \(r \gt 1\text{,}\) then \(r^N\) grows to \(\infty\) as \(N\rightarrow\infty\text{.}\)
  • If \(r \lt -1\text{,}\) then the magnitude of \(r^N\) grows to \(\infty\text{,}\) and the sign of \(r^N\) oscillates between \(+\) and \(-\text{,}\) as \(N\rightarrow\infty\text{.}\)
  • If \(r=+1\text{,}\) then \(N+1\) grows to \(\infty\) as \(N\rightarrow\infty\text{.}\)
  • If \(r=-1\text{,}\) then \(r^N\) just oscillates between \(+1\) and \(-1\) as \(N\rightarrow\infty\text{.}\)
In each case the sequence of partial sums does not converge and so the series does not converge.
Here are some sketches of the graphs of \(\frac{1}{1-r}\) and \(S_N\text{,}\) \(0\le N\le 5\text{,}\) for \(a=1\) and \(-1\le r\lt 1\text{.}\)
In these sketches we see that
  • when \(0\lt r \lt 1\text{,}\) and also when \(-1\lt r\lt 0\) with \(N\) odd, we have \(S_N=\frac{1-r^{N+1}}{1-r}\lt \frac{1}{1-r}\text{.}\) On the other hand, when \(-1\lt r\lt 0\) with \(N\) even, we have \(S_N=\frac{1-r^{N+1}}{1-r}\gt \frac{1}{1-r}\text{.}\)
  • When \(0\lt |r|\lt 1\text{,}\) \(S_N=\frac{1-r^{N+1}}{1-r}\) gets closer and closer to \(\frac{1}{1-r}\) as \(N\) increases.
  • When \(r=-1\text{,}\) \(S_N\) just alternates between \(0\text{,}\) when \(N\) is odd, and \(1\text{,}\) when \(N\) is even.
We should summarise the results in the previous example in a lemma.
Now that we know how to handle geometric series let’s return to Example 3.2.1.

Example 3.2.6. Decimal Expansions.

The decimal expansion
\begin{equation*} 0.3333\cdots =\frac{3}{10}+\frac{3}{100}+\frac{3}{1000}+\frac{3}{10000}+\cdots =\sum_{n=1}^\infty\frac{3}{10^n} \end{equation*}
is a geometric series with the first term \(a=\frac{3}{10}\) and the ratio \(r=\frac{1}{10}\text{.}\) So, by Lemma 3.2.5,
\begin{equation*} 0.3333\cdots =\sum_{n=1}^\infty\frac{3}{10^n} =\frac{\frac{3}{10}}{1-\frac{1}{10}} =\frac{\frac{3}{10}}{\frac{9}{10}} =\frac{1}{3} \end{equation*}
just as we would have expected.
We can push this idea further. Consider the repeating decimal expansion:
\begin{equation*} 0.16161616\cdots =\frac{16}{100}+\frac{16}{10000}+\frac{16}{1000000}+\cdots \end{equation*}
This is another geometric series with the first term \(a=\frac{16}{100}\) and the ratio \(r=\frac{1}{100}\text{.}\) So, by Lemma 3.2.5,
\begin{equation*} 0.16161616\cdots =\sum_{n=1}^\infty\frac{16}{100^n} =\frac{\frac{16}{100}}{1-\frac{1}{100}} =\frac{\frac{16}{100}}{\frac{99}{100}} =\frac{16}{99} \end{equation*}
again, as expected. In this way any periodic decimal expansion converges to a ratio of two integers — that is, to a rational number
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We have included a (more) formal proof of this fact in the optional §3.7 at the end of this chapter. Proving that a repeating decimal expansion gives a rational number isn’t too hard. Proving the converse — that every rational number has a repeating decimal expansion is a little trickier, but we also do that in the same optional section.
.
Here is another more complicated example.
\begin{align*} 0.1234343434\cdots &=\frac{12}{100}+\frac{34}{10000}+\frac{34}{1000000}+\cdots\\ &=\frac{12}{100}+\sum_{n=2}^\infty\frac{34}{100^n}\\ \end{align*}

Now apply Lemma 3.2.5 with \(a=\frac{34}{100^2}\) and \(r=\frac{1}{100}\)

\begin{align*} &=\frac{12}{100}+\frac{34}{10000}\frac{1}{1-\frac{1}{100}}\\ &=\frac{12}{100}+\frac{34}{10000}\frac{100}{99}\\ &=\frac{1222}{9900} \end{align*}
Typically, it is quite difficult to write down a neat closed form expression for the partial sums of a series. Geometric series are very notable exceptions to this. Another family of series for which we can write down partial sums is called “telescoping series”. These series have the desirable property that many of the terms in the sum cancel each other out rendering the partial sums quite simple.

Example 3.2.7. Telescoping Series.

In this example, we are going to study the series \(\sum_{n=1}^\infty\frac{1}{n(n+1)}\text{.}\) This is a rather artificial series
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Well… this sort of series does show up when you start to look at the Maclaurin polynomial of functions like \((1-x)\log(1-x)\text{.}\) So it is not totally artificial. At any rate, it illustrates the basic idea of telescoping very nicely, and the idea of “creative telescoping” turns out to be extremely useful in the study of series — though it is well beyond the scope of this course.
that has been rigged to illustrate a phenomenon called “telescoping”. Notice that the \(n^{\rm th}\) term can be rewritten as
\begin{align*} \frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}\\ \end{align*}

and so we have

\begin{align*} a_n &= b_n - b_{n+1} & \text{where } b_n &= \frac{1}{n}. \end{align*}
Because of this we get big cancellations when we add terms together. This allows us to get a simple formula for the partial sums of this series.
\begin{align*} S_N&= \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \cdots + \frac{1}{N\cdot (N+1)}\\ &= \Big(\frac{1}{1}-\frac{1}{2}\Big) + \Big(\frac{1}{2}-\frac{1}{3}\Big) + \Big(\frac{1}{3}-\frac{1}{4}\Big) + \cdots + \Big(\frac{1}{N}-\frac{1}{N+1}\Big) \end{align*}
The second term of each bracket exactly cancels the first term of the following bracket. So the sum “telescopes” leaving just
\begin{equation*} S_N = 1-\frac{1}{N+1} \end{equation*}
and we can now easily compute
\begin{equation*} \sum_{n=1}^\infty\frac{1}{n(n+1)} =\lim_{N\rightarrow\infty} S_N =\lim_{N\rightarrow\infty}\Big( 1-\frac{1}{N+1}\Big) =1 \end{equation*}
More generally, if we can write
\begin{align*} a_n &= b_n - b_{n+1}\\ \end{align*}

for some other known sequence \(b_n\text{,}\) then the series telescopes and we can compute partial sums using

\begin{align*} \sum_{n=1}^N a_n &= \sum_{n=1}^N (b_n - b_{n+1})\\ &= \sum_{n=1}^N b_n - \sum_{n=1}^N b_{n+1}\\ &= b_1 - b_{N+1}.\\ \end{align*}

and hence

\begin{align*} \sum_{n=1}^\infty a_n &= b_1 - \lim_{N\to\infty} b_{N+1} \end{align*}
provided this limit exists.Often \(\lim\limits_{N\to\infty} b_{N+1}=0\) and then \(\sum\limits_{n=1}^\infty a_n =b_1\text{.}\) But this does not always happen. Here is an example.

Example 3.2.8. A Divergent Telescoping Series.

In this example, we are going to study the series \(\sum\limits_{n=1}^\infty\log\big(1+\frac{1}{n}\big)\text{.}\) Let’s start by just writing out the first few terms.
\begin{alignat*}{1} \sum_{n=1}^\infty\log\Big(1+\frac{1}{n}\Big) &=\overbrace{\log\Big(1+\frac{1}{1}\Big)}^{n=1} +\overbrace{\log\Big(1+\frac{1}{2}\Big)}^{n=2} +\overbrace{\log\Big(1+\frac{1}{3}\Big)}^{n=3}\\ \amp\hskip1in +\overbrace{\log\Big(1+\frac{1}{4}\Big)}^{n=4} +\cdots\\ &=\log(2)+\log\Big(\frac{3}{2}\Big) +\log\Big(\frac{4}{3}\Big)\\ \amp\hskip1in +\log\Big(\frac{5}{4}\Big) + \cdots \end{alignat*}
This is pretty suggestive since
\begin{gather*} \log(2)+\log\Big(\frac{3}{2}\Big) +\log\Big(\frac{4}{3}\Big) +\log\Big(\frac{5}{4}\Big) =\log\Big(2\times\frac{3}{2}\times\frac{4}{3}\times \frac{5}{4}\Big) =\log(5) \end{gather*}
So let’s try using this idea to compute the partial sum \(S_N\text{:}\)
\begin{align*} S_N&=\sum_{n=1}^N\log\Big(1+\frac{1}{n}\Big)\\ &=\overbrace{\log\Big(1+\frac{1}{1}\Big)}^{n=1} +\overbrace{\log\Big(1+\frac{1}{2}\Big)}^{n=2} +\overbrace{\log\Big(1+\frac{1}{3}\Big)}^{n=3} + \cdots\\ \amp\hskip1in +\overbrace{\log\Big(1+\frac{1}{N-1}\Big)}^{n=N-1} +\overbrace{\log\Big(1+\frac{1}{N}\Big)}^{n=N}\\ &=\log(2) + \log\Big(\frac{3}{2}\Big) +\log\Big(\frac{4}{3}\Big) +\cdots+ \log\Big(\frac{N}{N-1}\Big) + \log\Big(\frac{N+1}{N}\Big)\\ &=\log\Big(2\times\frac{3}{2}\times\frac{4}{3}\times \cdots \times \frac{N}{N-1}\times \frac{N+1}{N}\Big)\\ &=\log(N+1) \end{align*}
Uh oh!
\begin{gather*} \lim_{N\rightarrow\infty} S_N =\lim_{N\rightarrow\infty}\log(N+1) =+\infty \end{gather*}
This telescoping series diverges! There is an important lesson here. Telescoping series can diverge. They do not always converge to \(b_1\text{.}\)
As was the case for limits, differentiation and antidifferentiation, we can compute more complicated series in terms of simpler ones by understanding how series interact with the usual operations of arithmetic. It is, perhaps, not so surprising that there are simple rules for addition and subtraction of series and for multiplication of a series by a constant. Unfortunately there are no simple general rules for computing products or ratios of series.

Example 3.2.10. \(\sum_{n=1}^\infty\Big( \frac{1}{7^n}+\frac{2}{n(n+1)}\Big)\).

As a simple example of how we use the arithmetic of series Theorem 3.2.9, consider
\begin{equation*} \sum_{n=1}^\infty\Big[\frac{1}{7^n}+\frac{2}{n(n+1)}\Big] \end{equation*}
We recognize that we know how to compute parts of this sum. We know that
\begin{equation*} \sum_{n=1}^\infty\frac{1}{7^n}=\frac{\frac{1}{7}}{1-\frac{1}{7}} =\frac{1}{6} \end{equation*}
because it is a geometric series (Example 3.2.4 and Lemma 3.2.5) with first term \(a=\frac{1}{7}\) and ratio \(r=\frac{1}{7}\text{.}\) And we know that
\begin{equation*} \sum_{n=1}^\infty\frac{1}{n(n+1)} =1 \end{equation*}
by Example 3.2.7. We can now use Theorem 3.2.9 to build the specified “complicated” series out of these two “simple” pieces.
\begin{align*} \sum_{n=1}^\infty\Big[\frac{1}{7^n}+\frac{2}{n(n+1)}\Big] &=\sum_{n=1}^\infty \frac{1}{7^n} +\sum_{n=1}^\infty\frac{2}{n(n+1)} & \text{by Theorem }\knowl{./knowl/xref/thm_SRseriesarith.html}{\text{3.2.9}}\text{(a)}\\ &= \sum_{n=1}^\infty\frac{1}{7^n} +2 \sum_{n=1}^\infty \frac{1}{n(n+1)} & \text{by Theorem }\knowl{./knowl/xref/thm_SRseriesarith.html}{\text{3.2.9}}\text{(b)}\\ &=\frac{1}{6}+2\cdot 1 =\frac{13}{6} \end{align*}

Example 3.2.11. Arithmetic Progressions.

An arithmetic progression is a sequence for which the difference between successive terms is a constant. That is, there are constants \(a_1\) and \(d\) such that the \(n^{\rm th}\) term in the sequence is \(a_n = a_1 + (n-1) d\text{.}\) For example, the sequence \(1\text{,}\) \(2\text{,}\) \(3\text{,}\) \(4\text{,}\) \(\cdots\) has \(a_1=1\text{,}\) \(d=1\) and \(a_n=n\text{.}\) We will now compute the partial sum \(S_N = \sum_{n=1}^N n\) of the corresponding series \(\sum_{n=1}^\infty n\text{.}\) To do so, we will use a trick that is often attributed to Carl Friedrich Gauss
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Carl Friedrich Gauss (1777–1855) was a German mathematician and physicist who ranks among the world’s most influential mathematicians. He is said to have reinvented this trick when he was in primary school. In fact, the trick was known many centuries before him. You can use your favourite search engine to learn more.
. The trick consists of writing \(S_N\) twice, once in forward order and once in reverse order
\begin{align*} S_N &= 1\ +\ \ \ \ \ \ 2\ \ \ \ \ \ +\ \ \ \ \ \ 3\ \ \ \ \ +\cdots+ N\\ S_N &= N +(N-1) + (N-2) +\cdots+1 \end{align*}
and then adding the two representations of \(S_N\) together.
\begin{align*} 2 S_N & = \overbrace{(N+1) + (N+1) + (N+1) + \cdots (N+1)}^{N\ {\rm terms}} \\ & = N(N+1) \\ \implies S_N & = \frac{N(N+1)}{2} \end{align*}
Using this and a little arithmetic, we can compute the partial sums for the series that corresponds to the general arithmetic progression \(a_n=a_1 + (n-1) d\text{.}\) (Of course, you can get the same result by applying the above trick to \(S_N = \sum_{n=1}^N \big\{a_1 + (n-1) d\big\}\text{.}\))
\begin{align*} S_N & = \sum_{n=1}^N \big\{a_1 + (n-1) d\big\} = \sum_{n=1}^N \big\{(a_1 -d) + n d\big\} \\ & = (a_1-d) \sum_{n=1}^N 1 + d\sum_{n=1}^N n = (a_1-d)N + d\tfrac{N(N+1)}{2} \\ &=\tfrac{N}{2}\Big\{a_1 + [a_1+d(N-1)] \Big\} \\ &= \tfrac{N}{2}\big\{\text{first term}+\text{last term}\big\} \end{align*}

Example 3.2.12. Power Sequence.

We are now going to consider partial sums for the series \(\sum_{n=1}^\infty n^k\text{,}\) for several values of the parameter \(k\text{.}\) In the last example, we found the partial sum
\begin{gather*} \sum_{n=1}^N n = \tfrac{1}{2}N(N+1) \end{gather*}
for the case \(k=1\text{.}\) In Theorem  1.1.6 we stated the partial sums
\begin{align*} \sum_{n=1}^N n^2 &= \tfrac{1}{6}N(N+1)(2N+1) \\ \sum_{n=1}^N n^3 &= \Big[\tfrac{1}{2}N(N+1)\Big]^2 \end{align*}
for the cases \(k=2\) and \(k=3\text{.}\) In that theorem, while we gave a strategy that could be used to prove those formulae, we did not actually give the proofs. We are now going to give the proofs, using a different strategy — a strategy that can be used to find the partial sums for any natural number \(k\text{.}\) As a warmup, we’ll start with \(k=1\text{,}\) even though we have already handled that case. The strategy is based on the observation that the telescoping sum
\begin{align*} \sum_{n=1}^N \big\{(n+1)^2 -n^2\big\} &= (N+1)^2 -1 = N^2+2N \end{align*}
On the other hand, the same sum
\begin{align*} \sum_{n=1}^N \big\{(n+1)^2 -n^2\big\} &= \sum_{n=1}^N \big\{2n+1\big\} = 2\left[\sum_{n=1}^N n\right] + \left[\sum_{n=1}^N 1\right] \\ &= 2\left[\sum_{n=1}^N n\right] + N \end{align*}
Equating the two formulae for \(\sum_{n=1}^N \big\{(n+1)^2 -n^2\big\}\) gives
\begin{align*} &N^2+2N = 2\left[\sum_{n=1}^N n\right] + N \\ \implies &\sum_{n=1}^N n = \tfrac{1}{2}\big[N^2+N\big] =\tfrac{1}{2} N (N+1) \end{align*}
as we found in Example 3.2.11. Now we move on to \(k=2\text{.}\) This time we use the telescoping sum
\begin{align*} \sum_{n=1}^N \big\{(n+1)^3 -n^3\big\} &= (N+1)^3 -1 = N^3 + 3N^2 +3N \end{align*}
On the other hand, the same sum
\begin{align*} \sum_{n=1}^N \big\{(n+1)^3 -n^3\big\} &= \sum_{n=1}^N \big\{3n^2 + 3n+1\big\} = 3\left[\sum_{n=1}^N n^2\right] + 3\left[\sum_{n=1}^N n\right] + \left[\sum_{n=1}^N 1\right] \\ &= 3\left[\sum_{n=1}^N n^2\right] + \tfrac{3}{2}N(N+1) + N \end{align*}
Equating the two formulae for \(\sum_{n=1}^N \big\{(n+1)^3 -n^3\big\}\) gives
\begin{align*} N^3 + 3N^2 +3N &= 3\left[\sum_{n=1}^N n^2\right] + \tfrac{3}{2}N^2 +\tfrac{5}{2}N\\ \implies \sum_{n=1}^N n^2 &= \tfrac{1}{3}\Big[N^3 + \tfrac{3}{2}N^2 +\tfrac{1}{2}N\Big] =\tfrac{1}{6}N\Big[2N^2+3N+1\Big]\\ &=\tfrac{1}{6}N(N+1)(2N+1) \end{align*}
as stated in Theorem 1.1.6. Finally, we get to \(k=3\text{.}\) This time we use the telescoping sum
\begin{align*} \sum_{n=1}^N \big\{(n+1)^4 -n^4\big\} &= (N+1)^4 -1 \end{align*}
On the other hand, the same sum
\begin{align*} \sum_{n=1}^N \big\{(n+1)^4 -n^4\big\} &= \sum_{n=1}^N \big\{4n^3 +6n^2+4n+1\big\} \\ &= 4\left[\sum_{n=1}^N n^3\right] + 6\left[\sum_{n=1}^N n^2\right] + 4\left[\sum_{n=1}^N n\right] + \left[\sum_{n=1}^N 1\right] \\ &= 4\left[\sum_{n=1}^N n^3\right] + N(N+1)(2N+1) +2N(N+1) + N \end{align*}
Equating the two formulae for \(\sum_{n=1}^N \big\{(n+1)^4 -n^4\big\}\) gives
\begin{align*} (N+1)^4 -1 &= 4\left[\sum_{n=1}^N n^3\right] + (N+1)(2N^2+3N) + N \\ \implies \sum_{n=1}^N n^3 &= \tfrac{1}{4}\Big[(N+1)^4-1-(N+1)(2N^3+3N) -N\Big] \\ &=\tfrac{1}{4}(N+1)\Big[(N+1)^3-(2N^2+3N) -1\Big] \\ &=\tfrac{1}{4}(N+1)\Big[N^3+N^2\Big] \\ &=\tfrac{1}{4}N^2(N+1)^2 \end{align*}
as stated in Theorem 1.1.6.

Exercises 3.2.2 Exercises

Exercises — Stage 1 .

1.
Write out the first five partial sums corresponding to the series \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}\text{.}\)
You don’t need to simplify the terms.
2.
Every student who comes to class brings their instructor cookies, and leaves them on the instructor’s desk. Let \(C_k\) be the total number of cookies on the instructor’s desk after the \(k\)th student comes.
If \(C_{11}=20\text{,}\) and \(C_{10}=17\text{,}\) how many cookies did the 11th student bring to class?
3.
Suppose the sequence of partial sums of the series \(\displaystyle\sum_{n=1}^\infty a_n\) is \(\{S_N\} = \left\{\dfrac{N}{N+1}\right\}\text{.}\)
  1. What is \(\{a_n\}\text{?}\)
  2. What is \(\lim\limits_{n \to \infty} a_n \text{?}\)
  3. Evaluate \(\displaystyle\sum_{n=1}^\infty a_n\text{.}\)
4.
Suppose the sequence of partial sums of the series \(\displaystyle\sum_{n=1}^\infty a_n\) is \(\{S_N\} = \left\{(-1)^N+\dfrac{1}{N}\right\}\text{.}\)
What is \(\{a_n\}\text{?}\)
5.
Let \(f(N)\) be a formula for the \(N\)th partial sum of \(\displaystyle\sum_{n=1}^\infty a_n\text{.}\) (That is, \(f(N)=S_N\text{.}\)) If \(f'(N) \lt 0\) for all \(N \gt 1\text{,}\) what does that say about \(a_n\text{?}\)

Exercise Group.

Questions 6 through 8 invite you to explore geometric sums in a geometric way. This is complementary to than the algebraic method discussed in the text.
6.
Suppose the triangle outlined in red in the picture below has area one.
  1. Express the combined area of the black triangles as a series, assuming the pattern continues forever.
  2. Evaluate the series using the picture (not the formula from your book).
7.
Suppose the square outlined in red in the picture below has area one.
  1. Express the combined area of the black squares as a series, assuming the pattern continues forever.
  2. Evaluate the series using the picture (not the formula from your book).
8.
In the style of Questions 6 and 7, draw a picture that represents \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^n}\) as an area.
9.
Evaluate \(\displaystyle\sum_{n=0}^{100}\frac{1}{5^n}\text{.}\)
10.
Every student who comes to class brings their instructor cookies, and leaves them on the instructor’s desk. Let \(C_k\) be the total number of cookies on the instructor’s desk after the \(k\)th student comes.
If \(C_{20}=53\text{,}\) and \(C_{10}=17\text{,}\) what does \(C_{20}-C_{10}=36\) represent?
11.
Evaluate \(\displaystyle\sum_{n=50}^{100}\frac{1}{5^n}\text{.}\) (Note the starting index.)
12.
  1. Starting on day \(d=1\text{,}\) every day you give your friend $\(\frac{1}{d+1}\text{,}\) and they give $\(\frac{1}{d}\) back to you. After a long time, how much money have you gained by this arrangement?
  2. Evaluate \(\displaystyle\sum_{d=1}^\infty \left(\frac{1}{d}-\frac{1}{(d+1)}\right)\text{.}\)
  3. Starting on day \(d=1\text{,}\) every day your friend gives you $\((d+1)\text{,}\) and they take $\((d+2)\) from you. After a long time, how much money have you gained by this arrangement?
  4. Evaluate \(\displaystyle\sum_{d=1}^\infty \left((d+1)-(d+2)\right)\text{.}\)
13.
Suppose \(\displaystyle\sum_{n=1}^\infty a_n = A\text{,}\) \(\displaystyle\sum_{n=1}^\infty b_n = B\text{,}\) and \(\displaystyle\sum_{n=1}^\infty c_n = C\text{.}\)
Evaluate \(\displaystyle\sum_{n=1}^\infty\left( a_n+b_n+c_{n+1}\right)\text{.}\)
14.
Suppose \(\displaystyle\sum_{n=1}^\infty a_n = A\text{,}\) \(\displaystyle\sum_{n=1}^\infty b_n = B \neq 0\text{,}\) and \(\displaystyle\sum_{n=1}^\infty c_n = C\text{.}\)
True or false: \(\displaystyle\sum_{n=1}^\infty\left(\dfrac{ a_n}{b_n}+c_{n}\right) = \frac{A}{B}+C\).

Exercises — Stage 2 .

15. (✳).
To what value does the series \(\displaystyle 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27}+ \frac{1}{81}+ \frac{1}{243} + \cdots\) converge?
16. (✳).
Evaluate \(\displaystyle \sum_{k=7}^{\infty} \frac{1}{8^k}\)
17. (✳).
Show that the series \(\displaystyle \sum_{k=1}^{\infty} \bigg( \frac{6}{k^2} - \frac{6}{(k+1)^2} \bigg)\) converges and find its limit.
18. (✳).
Find the sum of the convergent series \(\displaystyle\sum\limits_{n=3}^\infty \bigg( \!\cos\Big( \frac \pi n \Big) - \cos\Big( \frac \pi{n+1} \Big) \bigg)\text{.}\)
19. (✳).
The \(n^{\rm th}\) partial sum of a series \(\displaystyle \sum_{n=1}^{\infty}a_n \) is known to have the formula \(\displaystyle s_n = \frac{1+3n}{5+4n}\text{.}\)
  1. (a) Find an expression for \(a_n\text{,}\) valid for \(n\ge2\text{.}\)
  2. (b) Show that the series \(\displaystyle \sum_{n=1}^{\infty}a_n \) converges and find its value.
20. (✳).
Find the sum of the series \(\displaystyle\sum\limits_{n=2}^\infty \frac{3\cdot 4^{n+1}}{8\cdot 5^n}\text{.}\) Simplify your answer completely.
21. (✳).
Relate the number \(0.2\bar{3} = 0.233333\ldots\) to the sum of a geometric series, and use that to represent it as a rational number (a fraction or combination of fractions, with no decimals).
22. (✳).
Express \(2.656565\ldots\) as a rational number, i.e. in the form \(p/q\) where \(p\) and \(q\) are integers.
23. (✳).
Express the decimal \(0.\overline{321}=0.321321321\ldots\) as a fraction.
24. (✳).
Find the value of the convergent series
\begin{gather*} \sum_{n=2}^\infty \bigg( \frac{2^{n+1}}{3^n} + \frac1{2n-1} - \frac1{2n+1} \bigg) \end{gather*}
Simplify your answer completely.
25. (✳).
Evaluate
\begin{gather*} \sum_{n=1}^\infty\bigg[{\Big(\frac{1}{3}\Big)}^n + {\Big(-\frac{2}{5}\Big)}^{n-1}\bigg] \end{gather*}
26. (✳).
Find the sum of the series \(\displaystyle\sum_{n=0}^\infty\frac{1+3^{n+1}}{4^n}\text{.}\)
27.
Evaluate \(\displaystyle\sum_{n=5}^\infty \log\left(\frac{n-3}{n}\right)\text{.}\)
28.
Evaluate \(\displaystyle\sum_{n=2}^\infty \left(\frac{2}{n}-\frac{1}{n+1}-\frac{1}{n-1} \right)\text{.}\)

Exercises — Stage 3 .

29.
An infinitely long, flat cliff has stones hanging off it, attached to thin wire of negligible mass. Starting at position \(x=1\text{,}\) every metre (at position \(x\text{,}\) where \(x\) is some whole number) the stone has mass \(\dfrac{1}{4^x}\) kg and is hanging \(2^x\) metres below the top of the cliff.
30.
Find the combined volume of an infinite collection of spheres, where for each whole number \(n=1,\,2,\,3,\,\ldots\) there is exactly one sphere of radius \(\dfrac{1}{\pi^n}\text{.}\)
31.
Evaluate \(\displaystyle\sum_{n=3}^\infty \left(\frac{\sin^2 n}{2^n}+\frac{\cos^2(n+1)}{2^{n+1}} \right)\text{.}\)
32.
Suppose a series \(\displaystyle\sum_{n=1}^\infty a_n\) has sequence of partial sums \(\{S_N\}\text{,}\) and the series \(\displaystyle\sum_{N=1}^\infty S_N\) has sequence of partial sums \(\{\mathbb{S}_M \}=\left\{ \displaystyle\sum_{N=1}^M S_N\right\}\text{.}\)
If \(\mathbb{S}_M = \dfrac{M+1}{M}\text{,}\) what is \(a_n\text{?}\)
33.
Create a bullseye using the following method:
Starting with a red circle of area 1, divide the radius into thirds, creating two rings and a circle. Colour the middle ring blue.
Continue the pattern with the inside circle: divide its radius into thirds, and colour the middle ring blue.
Continue in this way indefinitely: dividing the radius of the innermost circle into thirds, creating two rings and another circle, and colouring the middle ring blue.
What is the area of the red portion?
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