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Subsection 3.1.1 Sequences

In the discussion above we used the term “sequence” without giving it a precise mathematical meaning. Let us rectify this now.

Definition 3.1.1

A sequence is a list of infinitely  1 For the more pedantic reader, here we mean a countably infinite list of numbers. The interested (pedantic or otherwise) reader should look up countable and uncountable sets. many numbers with a specified order. It is denoted

\begin{equation*} \big\{a_1,\ a_2,\ a_3,\ \cdots,\ a_n,\ \cdots\big\} \quad\text{or}\quad \big\{a_n\big\} \quad\text{or}\quad \big\{a_n\big\}_{n=1}^\infty \end{equation*}

We will often specify a sequence by writing it more explicitly, like

\begin{gather*} \Big\{ a_n = f(n) \Big\}_{n=1}^\infty \end{gather*}

where \(f(n)\) is some function from the natural numbers to the real numbers.

Here are three sequences.

\begin{align*} &\Big\{1,\ \frac{1}{2},\ \frac{1}{3},\ \cdots,\ \frac{1}{n},\ \cdots\Big\} &&\text{or} &&\Big\{a_n=\frac{1}{n}\Big\}_{n=1}^\infty\\ &\Big\{1,\ 2,\ 3,\ \cdots,\ n,\ \cdots\Big\} &&\text{or} &&\Big\{a_n=n\Big\}_{n=1}^\infty\\ &\Big\{1,\ -1,\ 1,\ -1,\ \cdots,\ (-1)^{n-1},\ \cdots\Big\} &&\text{or} &&\Big\{a_n=(-1)^{n-1}\Big\}_{n=1}^\infty \end{align*}

It is not necessary that there be a simple explicit formula for the \(n^{\rm th}\) term of a sequence. For example the decimal digits of \(\pi\) is a perfectly good sequence

\begin{equation*} \big\{3,\ 1,\ 4,\ 1,\ 5,\ 9,\ 2,\ 6,\ 5,\ 3,\ 5,\ 8,\ 9,\ 7,\ 9,\ 3,\ 2,\ 3,\ 8,\ 4,\ 6,\ 2,\ 6,\ 4,\ \cdots\ \big\} \end{equation*}

but there is no simple formula  2 There is, however, a remarkable result due to Bailey, Borwein and Plouffe that can be used to compute the \(n^{\rm th}\) binary digit of \(\pi\) (i.e. writing \(\pi\) in base 2 rather than base 10) without having to work out the preceding digits. for the \(n^{\rm th}\) digit.

Our primary concern with sequences will be the behaviour of \(a_n\) as \(n\) tends to infinity and, in particular, whether or not \(a_n\) “settles down” to some value as \(n\) tends to infinity.

Definition 3.1.3

A sequence \(\big\{a_n\big\}_{n=1}^\infty\) is said to converge to the limit \(A\) if \(a_n\) approaches \(A\) as \(n\) tends to infinity. If so, we write

\begin{equation*} \lim_{n\rightarrow\infty} a_n=A\qquad\hbox{or}\qquad a_n\rightarrow A\text{ as }n\rightarrow\infty \end{equation*}

A sequence is said to converge if it converges to some limit. Otherwise it is said to diverge.

The reader should immediately recognise the similarity with limits at infinity

\begin{gather*} \lim_{x \to \infty} f(x) = L \qquad\hbox{if}\qquad f(x) \to L \text{ as } x \to \infty \end{gather*}

Three of the four sequences in Example 3.1.2 diverge:

  • The sequence \(\big\{a_n=n\big\}_{n=1}^\infty\) diverges because \(a_n\) grows without bound, rather than approaching some finite value, as \(n\) tends to infinity.
  • The sequence \(\big\{a_n=(-1)^{n-1}\big\}_{n=1}^\infty\) diverges because \(a_n\) oscillates between \(+1\) and \(-1\) rather than approaching a single value as \(n\) tends to infinity.
  • The sequence of the decimal digits of \(\pi\) also diverges, though the proof that this is the case is a bit beyond us right now  3 If the digits of \(\pi\) were to converge, then \(\pi\) would have to be a rational number. The irrationality of \(\pi\) (that it cannot be written as a fraction) was first proved by Lambert in 1761. Niven's 1947 proof is more accessible and we invite the interested reader to use their favourite search engine to find step-by-step guides to that proof. .

The other sequence in Example 3.1.2 has \(a_n=\frac{1}{n}\text{.}\) As \(n\) tends to infinity, \(\frac{1}{n}\) tends to zero. So

\begin{equation*} \lim_{n\rightarrow\infty} \frac{1}{n}=0 \end{equation*}

Here is a little less trivial example. To study the behaviour of \(\frac{n}{2n+1}\) as \(n\rightarrow\infty\text{,}\) it is a good idea to write it as

\begin{equation*} \frac{n}{2n+1}=\frac{1}{2+\frac{1}{n}} \end{equation*}

As \(n\rightarrow\infty\text{,}\) the \(\frac{1}{n}\) in the denominator tends to zero, so that the denominator \(2+\frac{1}{n}\) tends to \(2\) and \(\frac{1}{2+\frac{1}{n}}\) tends to \(\frac{1}{2}\text{.}\) So

\begin{gather*} \lim_{n\rightarrow\infty}\frac{n}{2n+1} =\lim_{n\rightarrow\infty}\frac{1}{2+\frac{1}{n}} =\frac{1}{2} \end{gather*}

Notice that in this last example, we are really using techniques that we used before to study infinite limits like \(\ds \lim_{x\rightarrow\infty}f(x)\text{.}\) This experience can be easily transferred to dealing with \(\lim\limits_{n\rightarrow\infty}a_n\) limits by using the following result.

Set \(f(x)=e^{-x}\text{.}\) Then \(e^{-n}=f(n)\) and

\begin{align*} \text{since } \lim_{x\rightarrow\infty}e^{-x}&=0 &\text{ we know that }&& \lim\limits_{n\rightarrow\infty}e^{-n}&=0 \end{align*}

The bulk of the rules for the arithmetic of limits of functions that you already know also apply to the limits of sequences. That is, the rules you learned to work with limits such as \(\ds \lim_{x\rightarrow\infty}f(x)\) also apply to limits like \(\ds\lim_{n\rightarrow\infty}a_n\text{.}\)

We use these rules to evaluate limits of more complicated sequences in terms of the limits of simpler sequences — just as we did for limits of functions.

Combining Examples 3.1.5 and 3.1.7,

\begin{align*} \lim_{n\rightarrow\infty}\Big[\frac{n}{2n+1} + 7 e^{-n}\Big] &= \lim_{n\rightarrow\infty}\frac{n}{2n+1} +\lim_{n\rightarrow\infty} 7 e^{-n} & \text{by Theorem }\knowl{./knowl/thm_SRlimarith.html}{\text{3.1.8}}\text{(a)}\\ &= \lim_{n\rightarrow\infty}\frac{n}{2n+1} +7\lim_{n\rightarrow\infty} e^{-n} & \text{by Theorem }\knowl{./knowl/thm_SRlimarith.html}{\text{3.1.8}}\text{(c)}\\ \end{align*}

and then using Examples 3.1.5 and 3.1.7

\begin{align*} &=\frac{1}{2} + 7\cdot 0\\ &=\frac{1}{2} \end{align*}

There is also a squeeze theorem for sequences.

In this example we use the squeeze theorem to evaluate

\begin{equation*} \lim_{n\rightarrow\infty}\Big[1+\frac{\pi_n}{n}\Big] \end{equation*}

where \(\pi_n\) is the \(n^{\mathrm{th}}\) decimal digit of \(\pi\text{.}\) That is,

\begin{equation*} \pi_1=3\quad \pi_2=1 \quad \pi_3=4 \quad \pi_4=1 \quad \pi_5=5 \quad\pi_6=9\quad\cdots \end{equation*}

We do not have a simple formula for \(\pi_n\text{.}\) But we do know that

\begin{equation*} 0\le\pi_n\le 9 \implies 0 \le \frac{\pi_n}{n} \le \frac{9}{n} \implies 1 \le 1+\frac{\pi_n}{n} \le 1+\frac{9}{n} \end{equation*}

and we also know that

\begin{gather*} \lim_{n\rightarrow\infty} 1 = 1\qquad \lim_{n\rightarrow\infty} \Big[1+\frac{9}{n}\Big] = 1 \end{gather*}

So the squeeze theorem with \(a_n=1\text{,}\) \(b_n=1+\frac{\pi_n}{n}\text{,}\) and \(c_n=1+\frac{9}{n}\) gives

\begin{equation*} \lim_{n\rightarrow\infty}\Big[1+\frac{\pi_n}{n}\Big] = 1 \end{equation*}

Finally, recall that we can compute the limit of the composition of two functions using continuity. In the same way, we have the following result:

Write \(\sin\frac{\pi n}{2n+1}=g\big(\frac{n}{2n+1}\big)\) with \(g(x)=\sin(\pi x)\text{.}\) We saw, in Example 3.1.5 that

\begin{equation*} \lim_{n\rightarrow\infty}\frac{n}{2n+1} = \frac{1}{2} \end{equation*}

Since \(g(x) = \sin (\pi x)\) is continuous at \(x=\frac{1}{2}\text{,}\) which is the limit of \(\frac{n}{2n+1}\text{,}\) we have

\begin{equation*} \lim_{n\rightarrow\infty}\sin\frac{\pi n}{2n+1} =\lim_{n\rightarrow\infty}g\Big(\frac{n}{2n+1}\Big) =g\Big(\frac{1}{2}\Big) =\sin\frac{\pi}{2} =1 \end{equation*}

With this introduction to sequences and some tools to determine their limits, we can now return to the problem of understanding infinite sums.

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