3.1 Infinite series of real numbers: definition and basic examples

Using the definition of a limit of a sequence from the previous section, we can introduce the concept of a convergent series.

Definition 3.1.

Let $(a_{k})_{k\in\mathbb{N}}$ be a sequence of real numbers.

1

The sequence of partial sums $(s_{n})_{n\in\mathbb{N}}$ is given by

$s_{n}:=\sum_{k=1}^{n}a_{k}\qquad\text{for all $n\in\mathbb{N}$.}$
2

We say that the series $\sum_{k=1}^{\infty}a_{k}$ converges (respectively, diverges) if the sequence of partial sums $(s_{n})_{n\in\mathbb{N}}$ converges (respectively, diverges).
3

If the series $\sum_{k=1}^{\infty}a_{k}$ converges, then we let $\sum_{k=1}^{\infty}a_{k}$ denote the value of the limit $\lim_{n\to\infty}s_{n}$ , which we refer to as the limit of the series.

Warning 3.2.

Once again, here ‘ $\infty$ ’ is just a symbol and does not represent a mathematical object. Despite the suggestive notation, one should avoid thinking of $\sum_{k=1}^{\infty}a_{k}$ as ‘summing together infinitely many terms’: in the definition, we always sum together only finite terms (indeed, summation only makes sense for finitely many terms) and then take a limit of the finite sums.¹¹ 1 When we study the Riemann rearrangement theorem later in this chapter, we shall see that series $\sum_{k=1}^{\infty}a_{k}$ can behave very differently compared to the bona fide (finite) sums $\sum_{k=1}^{n}a_{k}$ .

Geometric series

In some rare instances, it is possible to derive a compact formula for the partial sums $(s_{n})_{n\in\mathbb{N}}$ of a series, which can then be used to determine the convergence properties. One important example of this is given by geometric series.

Definition 3.3.

Given $a$ , $r\in\mathbb{R}$ , we say $\sum_{k=1}^{\infty}ar^{k}$ is a geometric series.

For any $a$ , $r\in\mathbb{R}$ with $r\neq 1$ , there is a simple formula for the partial sums $(s_{n})_{n\in\mathbb{N}}$ of the geometric series $\sum_{k=1}^{\infty}ar^{k}$ . To derive this formula, we fix $n\in\mathbb{N}$ and consider $s_{n}$ and $rs_{n}$ . Writing out these expressions, we see

$\displaystyle s_{n}$	$\displaystyle=ar+ar^{2}+ar^{3}+\cdots+ar^{n},$
$\displaystyle rs_{n}$	$\displaystyle=\phantom{ar+{}}ar^{2}+ar^{3}+\cdots+ar^{n}+ar^{n+1}.$

If we form the difference then all terms cancel except for the first term in $s_{n}$ and the last term in $rs_{n}$ . Thus, we have $s_{n}-rs_{n}=ar-ar^{n+1}$ . If $r\neq 1$ , then this rearranges to give

(3.1) (3.1)

s_{n}=\frac{ar-ar^{n+1}}{1-r}=\frac{ar(1-r^{n})}{1-r}.

Recall from Exercise 2.31 that the sequence $(r^{n})_{n\in\mathbb{N}}$ converges to $0$ for $|r|<1$ and diverges for $|r|>1$ . Combining the above observations, we arrive at the following.

Example 3.4 (Geometric series).

Given $r\in\mathbb{R}$ we have

(3.2) (3.2)

\sum_{k=1}^{\infty}ar^{k}=\frac{ar}{1-r}\quad\text{for $|r|<1$}\qquad\text{and% }\qquad\sum_{k=1}^{\infty}r^{k}\quad\text{diverges for $|r|>1$.}

As an explicit example, taking $a=1$ and $r=1/2$ , we deduce that $\sum_{k=1}^{\infty}2^{-k}=1$ .

Exercise 3.5.

This exercise further explores the formula (3.1).

(i)

Rewrite the proof of (3.1) this time using $\Sigma$ notation for the sums.
(ii)

Give a new proof of (3.1) using induction.

Note that Example 3.4 does not treat the cases $r=1$ or $r=-1$ , which can be dealt with separately. For simplicity, in what follows we take $a=1$ .

Example 3.6.

The series $\sum_{k=1}^{\infty}1$ diverges.

Proof.

A direct computation shows that the sequence of partial sums $(s_{n})_{n\in\mathbb{N}}$ is given by $s_{n}=n$ for all $n\in\mathbb{N}$ . By the boundedness test for sequences, the sequence of partial sums $(n)_{n\in\mathbb{N}}$ diverges. Thus, the series $\sum_{k=1}^{\infty}1$ diverges. ∎

Example 3.7.

The series $\sum_{k=1}^{\infty}(-1)^{k}$ diverges.

Proof.

A direct computation shows that the sequence of partial sums $(s_{n})_{n\in\mathbb{N}}$ is given by $s_{n}=-1$ if $n$ is odd and $s_{n}=0$ if $n$ is even. By the subsequence test, the sequence of partial sums $(s_{n})_{n\in\mathbb{N}}$ diverges. Thus, the series $\sum_{k=1}^{\infty}(-1)^{k}$ diverges. ∎

Decimal expansions

We use series to give a precise definition of infinite decimal expansions.

Definition 3.8 (Decimal expansion).

Given $x\in[0,1)$ and a sequence $(d_{k})_{k\in\mathbb{N}}$ with $d_{k}\in\{0,1,\dots,9\}$ for all $k\in\mathbb{N}$ , we write $x=0.d_{1}d_{2}d_{3}\ldots$ if

x=\sum_{k=1}^{\infty}\frac{d_{k}}{10^{k}}.

In Definition 3.8 we only consider real numbers in $[0,1)$ , but one can derive the expansion for a general number $x\in\mathbb{R}$ from this by including the integer part.

Example 3.9.

The geometric series formula (3.2) provides some explicit examples of infinite decimal expansions. For instance,

0.55555\dots=\sum_{k=1}^{\infty}\frac{5}{10^{k}}=\frac{5\cdot 10^{-1}}{1-10^{-% 1}}=\frac{5}{9}.

You studied methods for computing decimal expansions of rational numbers based on this formula in IMU.

The following theorem tells us that every real number $x\in[0,1)$ can be expressed in terms of decimal digits.

Theorem 3.10 (Decimal expansion).

Let $x\in[0,1)$ . There exists a sequence $(d_{k})_{k\in\mathbb{N}}$ with $d_{k}\in\{0,1,\dots,9\}$ for all $k\in\mathbb{N}$ such that $x=\sum_{k=1}^{\infty}x_{k}10^{-k}$ . In other words, $x=0.d_{1}d_{2}d_{3}\ldots$ .

The idea behind Theorem 3.10 is very simple. We start with $x\in[0,1)$ , which we imagine as some point on the unit interval in the number line. Divide the unit interval $[0,1)$ into ten equal intervals

\Big{[}0,\,\frac{1}{10}\Big{)},\quad\Big{[}\frac{1}{10},\,\frac{2}{10}\Big{)},% \quad\Big{[}\frac{2}{10},\,\frac{3}{10}\Big{)},\quad\dots,\quad\Big{[}\frac{9}% {10},\,1\Big{)}.

Our point $x$ must lie in one of these intervals, say $x\in[1/10,2/10)$ . In this case we set $d_{1}=1$ and note that $d_{1}\cdot 10^{-1}\leq x<d_{1}\cdot 10^{-1}+10^{-1}$ , so the fraction $0.d_{1}$ is within $1/10$ of $x$ . We now split $[1/10,2/10)$ into ten equal intervals

\Big{[}\frac{1}{10},\,\frac{11}{100}\Big{)},\quad\Big{[}\frac{11}{10},\,\frac{% 12}{100}\Big{)},\quad\Big{[}\frac{12}{100},\,\frac{13}{100}\Big{)},\quad\dots,% \quad\Big{[}\frac{19}{100},\,\frac{2}{10}\Big{)}.

Again, our point $x$ must lie in one of these intervals, say $x\in[14/100,15/100)$ . In this case we set $d_{2}=4$ and note that

d_{1}\cdot 10^{-1}+d_{2}\cdot 10^{-2}\leq x<d_{1}\cdot 10^{-1}+d_{2}\cdot 10^{% -2}+10^{-2},

so the fraction $0.d_{1}d_{2}$ is within $1/100$ of $x$ . See Figure 3.1. We can continue in this way, obtaining better and better decimal approximants to $x$ which, in the limit, converge to $x$ .

(a) Since

x\in\big{[}\tfrac{1}{10},\tfrac{2}{10}\big{)}

, we set

d_{1}:=1

(b) Since

x\in\big{[}\tfrac{14}{100},\tfrac{15}{100}\big{)}

, we set

d_{2}:=4

Figure 3.1: Finding the decimal expansion of

x\in[0,1]

. The right-hand figure is a magnification of the red interval in the left-hand figure.

Whilst the idea is simple to state, working out the precise details of how this works in general is a little involved.

Proof (of Theorem 3.10).

Let $x\in[0,1)$ . We recursively construct a sequence of integers $d_{k}\in\{0,1,\dots,9\}$ such that

s_{n}\leq x<s_{n}+10^{-n}\qquad\text{for all $n\in\mathbb{N}$}\qquad\text{% where}\qquad s_{n}:=\sum_{k=1}^{n}d_{k}10^{-k}.

Once we have such a sequence, it is clear that $\sum_{k=1}^{\infty}d_{k}10^{-k}=x$ as a consequence of the squeeze theorem.

To start the sequence, observe there exists a unique $d_{1}\in\{0,1,\dots,9\}$ such that

\frac{d_{1}}{10}\leq x<\frac{d_{1}+1}{10}.

Then $s_{1}=d_{1}10^{-1}$ satisfies $s_{1}\leq x<s_{1}+10^{-1}$ , as required.

Suppose $d_{1},\dots,d_{n}\in\{0,1,\dots,9\}$ have been constructed and satisfy the desired properties. In particular, $0\leq x-s_{n}<10^{-n}$ so that $0\leq 10^{n}(x-s_{n})<1$ . Thus, there exists a unique value of $d_{n+1}\in\{0,1,\dots,9\}$ such that

\frac{d_{n+1}}{10}\leq 10^{n}(x-s_{n})<\frac{d_{n+1}+1}{10}.

Rearranging, we see that

s_{n+1}=s_{n}+d_{n+1}\cdot 10^{-(n+1)}\leq s_{n}+x-s_{n}=x<s_{n}+(d_{n+1}+1)% \cdot 10^{-n-1}=s_{n+1}+10^{-(n+1)},

as required. ∎

Remark 3.11.

One unfortunate property of decimal expansions is that they are, in general, not unique. For instance, we know two distinct expansions for the number $1$ : that is, $1=1.000\ldots$ and $1=0.999\ldots$ . Because of this non-uniqueness, whilst decimal expansions are useful for representing real numbers, they are a messy tool for defining real numbers. This is one reason why it is helpful to define the real numbers using the completeness axiom.

Remark 3.12.

There is nothing special about the decimal expansion, involving powers of $10$ , in the above. We can use the same ideas to study binary expansions $\sum_{k=1}^{\infty}d_{k}\cdot 2^{-k}$ where $d_{k}\in\{0,1\}$ , ternary expansions $\sum_{k=1}^{\infty}d_{k}\cdot 3^{-k}$ where $d_{k}\in\{0,1,2\}$ , and so on.

Telescoping series

We consider another example where it is possible to derive a compact formula for the partial sums.

Definition 3.13 (Telescoping series).

We say $\sum_{k=1}^{\infty}a_{k}$ is a telescoping series if there exists a sequence $(b_{k})_{k\in\mathbb{N}}$ such that $a_{k}=b_{k}-b_{k+1}$ for all $k\in\mathbb{N}$ .

For a telescoping series $\sum_{k=1}^{\infty}a_{k}$ with $a_{k}=b_{k}-b_{k+1}$ for all $k\in\mathbb{N}$ , we can use cancellation between terms to evaluate the partial sums explicitly in terms of $(b_{k})_{k\in\mathbb{N}}$ . In particular,

$\displaystyle\sum_{k=1}^{n}a_{k}$	$\displaystyle=a_{1}+a_{2}+\cdots+a_{n}$
	$\displaystyle=b_{1}-b_{2}+b_{2}-b_{3}+\cdots+b_{n}-b_{n+1}$
(3.3)	(3.3)	$\displaystyle=b_{1}-b_{n+1}.$

This means that the convergence of $\sum_{k=1}^{\infty}a_{k}$ is equivalent to the convergence of $(b_{n})_{n\in\mathbb{N}}$ .

Example 3.14.

We show that $\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1$ .

Proof.

By writing $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ , we see that $\sum_{k=1}^{\infty}\frac{1}{k(k+1)}$ is a telescoping series (with $b_{k}=\frac{1}{k}$ ). Using cancellation between terms as in (3.1), the $n$ th partial sum is given by

s_{n}:=\sum_{k=1}^{n}\frac{1}{k(k+1)}=1-\frac{1}{n+1}\qquad\text{for all $n\in% \mathbb{N}$.}

The terms on the right-hand side form a convergent sequence with limit equal to $1$ and so $s_{n}\to 1$ as $n\to\infty$ . Thus, by definition, $\sum_{k=1}^{\infty}\frac{1}{k(k+1)}=1$ . ∎

Exercise 3.15.

Show that $\sum_{k=1}^{\infty}\frac{1}{k^{2}+3k+2}=\frac{1}{2}$ . Hint: rewrite the terms using the partial fraction decomposition.

Example 3.16.

The series $\sum_{k=1}^{\infty}(\sqrt{k+1}-\sqrt{k})$ diverges.

Proof.

Writing $\sqrt{k+1}-\sqrt{k}=-\sqrt{k}-(-\sqrt{k+1})$ , this series is clearly a telescoping series. Using cancellation between terms as in (3.1), the $n$ th partial sum is given by

s_{n}:=\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=\sqrt{n+1}-1\qquad\text{for all $n% \in\mathbb{N}$.}

The terms on the right-hand side form an unbounded sequence and so $(s_{n})_{n\in\mathbb{N}}$ diverges by the boundedness test. Thus, by definition, $\sum_{k=1}^{\infty}(\sqrt{k+1}-\sqrt{k})$ diverges. ∎

Exercise 3.17.

Let $r\in\mathbb{R}$ be such that $|r|>1$ . By writing $r^{k}=r^{k}\frac{r-1}{r-1}$ , show that $\sum_{k=1}^{\infty}r^{k}$ is a telescoping series which diverges.