Series

Lessons

Senior 6 · Mathematics

Series

4 lessons

Lesson 2 of 4

Infinite series and tests of convergence

25 min

Can infinitely many numbers add up to something finite?

Walk halfway to a wall, then half the remaining distance, then half of what remains, and so on. You take infinitely many steps, of lengths $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots$ of the distance, yet you never pass the wall, and you get as close to it as you like. So it is reasonable to say that $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1$ . But not every infinite sum settles down: $1 + 1 + 1 + \cdots$ clearly does not. This lesson makes "adding infinitely many terms" precise and gives tests that decide which series have a sum.

Each piece is half of what is left, so the pieces never overflow the square, and they cover all of it: $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1$ .

Definition

Partial sum of the infinite series $\sum_{r=1}^{\infty} u_r$ : the finite sum $S_n = u_1 + u_2 + \cdots + u_n$ .

Convergent series: the series converges to the sum $S$ if the sequence of partial sums has the finite limit $S$ :

\sum_{r=1}^{\infty} u_r = S \iff \lim_{n \to +\infty} S_n = S .

Otherwise the series diverges: $S_{n}$ tends to $\pm\infty$ , or has no limit at all. Example: $\sum_{r=1}^{\infty} 1$ diverges, because $S_n = n \to +\infty$ ; $\sum_{r=1}^{\infty}(-1)^{r}$ diverges, because $S_{n}$ takes the values $-1, 0, -1, 0, \ldots$ and has no limit.

Important

Geometric series. For $a \ne 0$ , the series $\sum_{k=1}^{\infty} ar^{k-1} = a + ar + ar^2 + \cdots$ (the index is $k$ here, because $r$ is the ratio)

converges if $|r| \lt 1$ , with sum to infinity $S_\infty = \dfrac{a}{1 - r}$ ;
diverges if $|r| \ge 1$ .

Proof: $S_n = \frac{a(1 - r^n)}{1 - r}$ for $r \ne 1$ . If $|r| \lt 1$ , then $r^n \to 0$ (S5, Convergent and divergent sequences), so $S_n \to \frac{a}{1 - r}$ . If $|r| \gt 1$ , $|r^n| \to +\infty$ ; if $r = 1$ , $S_{n} = n a$ ; if $r = - 1$ , $S_{n}$ alternates between $a$ and $0$ . In none of these cases does $S_{n}$ have a finite limit.

Partial sums of $\sum_{r=1}^{\infty}\left(\frac{1}{2}\right)^{r-1}$ (blue) and $\sum_{r=1}^{\infty}\left(-\frac{1}{2}\right)^{r-1}$ (orange), computed exactly. Both converge: $\frac{1}{1 - \frac{1}{2}} = 2$ and $\frac{1}{1 + \frac{1}{2}} = \frac{2}{3}$ .

Worked example 1: geometric series

(a) Write $0.272727\ldots$ as a fraction. (b) Find $\sum_{r=1}^{\infty} 3\left(-\frac{1}{2}\right)^{r-1}$ . (c) For which $x$ does $\sum_{r=0}^{\infty}(2x - 1)^r$ converge, and what is its sum?

Solution

$0.2727\ldots = \frac{27}{100} + \frac{27}{100^2} + \cdots$ , with $a = \frac{27}{100}$ and $r = \frac{1}{100}$ : $S_\infty = \frac{\frac{27}{100}}{1 - \frac{1}{100}} = \frac{27}{99} = \frac{3}{11}$ .
$a = 3$ , $r = -\frac{1}{2}$ , $|r| \lt 1$ : $S_\infty = \frac{3}{1 + \frac{1}{2}} = 2$ .
The ratio is $2 x - 1$ , so the series converges if and only if $|2x - 1| \lt 1$ , that is, $0 \lt x \lt 1$ . Then the sum is $\frac{1}{1 - (2x - 1)} = \frac{1}{2 - 2x}$ .

Important

The $n$ th-term test (test for divergence). If $\sum u_r$ converges, then $u_r \to 0$ . So if $u_{r}$ does not tend to 0, the series diverges.

Proof: $u_n = S_n - S_{n-1} \to S - S = 0$ .

Example: $\sum \frac{r}{r + 1}$ diverges, since $\frac{r}{r + 1} \to 1 \ne 0$ .

The harmonic series: terms that shrink, a sum that does not

The harmonic series $\sum_{r=1}^{\infty}\frac{1}{r} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots$ has terms tending to 0, yet it diverges. Group the terms in blocks that end at powers of 2 and replace each term by the smallest in its block:

\begin{aligned} & 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \cdots + \frac{1}{8}\right) + \cdots \\ \ge{} & 1 + \frac{1}{2} + \frac{2}{4} + \frac{4}{8} + \cdots \end{aligned}

Each block adds at least $\frac{1}{2}$ , so $S_{2^k} \ge 1 + \frac{k}{2}$ , which exceeds any number when $k$ is large. The growth is very slow: $S_{1024} \approx 7.51$ , just above the bound $1 + \frac{10}{2} = 6$ . This argument dates from the fourteenth century and is due to Nicole Oresme.

Important

Comparison test (series with non-negative terms). Suppose $0 \le u_r \le v_r$ for all $r$ .

If $\sum v_r$ converges, so does $\sum u_r$ .
If $\sum u_r$ diverges, so does $\sum v_r$ .

Why: the partial sums of a series of non-negative terms increase, and an increasing sequence that is bounded above converges. If $\sum v_r = V$ , every partial sum of $\sum u_r$ is at most $V$ .

Series to compare with: geometric series, and $\sum \frac{1}{r^p}$ , which converges if $p \gt 1$ and diverges if $p \le 1$ . For $p = 2$ : $\frac{1}{r^2} \lt \frac{1}{r(r - 1)} = \frac{1}{r - 1} - \frac{1}{r}$ for $r \ge 2$ , so $\sum_{r=2}^{n}\frac{1}{r^2} \lt 1 - \frac{1}{n} \lt 1$ , and $\sum \frac{1}{r^2}$ converges. For $p \le 1$ , $\frac{1}{r^p} \ge \frac{1}{r}$ , so the series diverges like the harmonic series.

Worked example 2: comparison

Decide whether each series converges: (a) $\sum_{r=1}^{\infty}\frac{1}{r^2 + 3}$ (b) $\sum_{r=1}^{\infty}\frac{1}{\sqrt{r}}$ (c) $\sum_{r=1}^{\infty}\frac{2r + 1}{r^2 + r}$ .

Solution

$0 \lt \frac{1}{r^2 + 3} \lt \frac{1}{r^2}$ , and $\sum\frac{1}{r^2}$ converges, so the series converges.
$\frac{1}{\sqrt{r}} \ge \frac{1}{r}$ for $r \ge 1$ , and the harmonic series diverges, so the series diverges.
$\frac{2r + 1}{r(r + 1)} - \frac{1}{r} = \frac{2r + 1 - (r + 1)}{r(r + 1)} = \frac{1}{r + 1} \gt 0$ , so the terms exceed $\frac{1}{r}$ and the series diverges.

Important

Ratio test (d'Alembert). For a series of positive terms, let $L = \lim_{r \to +\infty}\dfrac{u_{r+1}}{u_r}$ .

If $L \lt 1$ , the series converges.
If $L \gt 1$ (or $L = +\infty$ ), it diverges.
If $L = 1$ , the test gives no decision.

Why: if $L \lt 1$ , choose $q$ with $L \lt q \lt 1$ ; from some term on, each term is less than $q$ times the one before, so the tail is smaller than a convergent geometric series. If $L \gt 1$ , the terms eventually grow, so they do not tend to 0.

Worked example 3: the ratio test

(a) $\sum_{r=1}^{\infty}\frac{r}{2^r}$ : $\frac{u_{r+1}}{u_r} = \frac{r + 1}{2^{r+1}}\cdot\frac{2^r}{r} = \frac{r + 1}{2r} \to \frac{1}{2} \lt 1$ : converges (its sum is 2).

(b) $\sum_{r=1}^{\infty}\frac{3^r}{r!}$ : $\frac{3^{r+1}}{(r + 1)!}\cdot\frac{r!}{3^r} = \frac{3}{r + 1} \to 0$ : converges.

(c) $\sum_{r=1}^{\infty}\frac{r!}{10^r}$ : $\frac{(r + 1)!}{10^{r+1}}\cdot\frac{10^r}{r!} = \frac{r + 1}{10} \to +\infty$ : diverges.

Important

Alternating series test (Leibniz). If $u_{r}$ decreases and $u_r \to 0$ , then $\sum_{r=1}^{\infty}(-1)^{r+1}u_r = u_1 - u_2 + u_3 - \cdots$ converges, and the error in stopping after $n$ terms is at most the first term left out: $|S - S_n| \le u_{n+1}$ .

Example: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ converges (Lesson 3 shows its sum is $\ln 2$ ), although the series of absolute values, the harmonic series, diverges. A series is absolutely convergent if $\sum |u_r|$ converges; an absolutely convergent series is convergent.

Warning

$u_r \to 0$ is necessary but not enough. The harmonic series has $\frac{1}{r} \to 0$ and still diverges. The $n$ th-term test can only prove divergence.
$L = 1$ in the ratio test decides nothing: both $\sum\frac{1}{r}$ (divergent) and $\sum\frac{1}{r^2}$ (convergent) give $L = 1$ . Use a comparison instead.

Tip

A good order of attack: (1) do the terms tend to 0? If not, it diverges. (2) Is it geometric, or like $\frac{1}{r^p}$ ? (3) Compare with one of those. (4) Factorials or powers like $2^{r}$ : ratio test. (5) Signs that alternate: Leibniz.

Eddie Woo: Divergence of the Harmonic Series (Exam Question 12 of 12). Watch an exam-style proof that the harmonic series diverges, and compare it with the grouping argument above.

✦

Practice

Name the test you use in each part.

Find the sum to infinity: (a) $12 + 4 + \frac{4}{3} + \cdots$ (b) $5 - 2 + \frac{4}{5} - \cdots$ .
Write $0.454545\ldots$ as a fraction in its lowest terms.
For which real $x$ does $\sum_{r=0}^{\infty}\left(\frac{x}{3}\right)^r$ converge? Find its sum.
Show that $\sum_{r=1}^{\infty}\frac{r}{r + 1}$ diverges.
Decide whether each series converges: (a) $\sum_{r=1}^{\infty}\frac{1}{r^2 + r}$ (b) $\sum_{r=3}^{\infty}\frac{\ln r}{r}$ .
Use the ratio test: (a) $\sum\frac{r^2}{3^r}$ (b) $\sum\frac{2^r}{r^2}$ (c) $\sum\frac{r!}{r^r}$ .
Show that $\sum_{r=1}^{\infty}\frac{(-1)^{r+1}}{\sqrt{r}}$ converges but is not absolutely convergent.
A ball is dropped from a height of 2 m, and after each bounce rises to $\frac{3}{5}$ of its previous height. Find the total distance it travels.

Show answersHide answers

(a) $r = \frac{1}{3}$ : $\frac{12}{1 - \frac{1}{3}} = 18$ (b) $r = -\frac{2}{5}$ : $\frac{5}{1 + \frac{2}{5}} = \frac{25}{7}$ .
$\frac{45}{99} = \frac{5}{11}$ .
$|x| \lt 3$ ; sum $\frac{1}{1 - \frac{x}{3}} = \frac{3}{3 - x}$ .
$n$ th-term test: $\frac{r}{r + 1} \to 1 \ne 0$ .
(a) Converges: $\frac{1}{r^2 + r} \lt \frac{1}{r^2}$ (in fact it telescopes to 1). (b) Diverges: for $r \ge 3$ , $\ln r \gt 1$ , so $\frac{\ln r}{r} \gt \frac{1}{r}$ .
(a) ratio $\frac{(r + 1)^2}{3r^2} \to \frac{1}{3}$ : converges (b) ratio $\frac{2r^2}{(r + 1)^2} \to 2$ : diverges (c) ratio $\frac{(r + 1)!}{(r + 1)^{r+1}}\cdot\frac{r^r}{r!} = \left(\frac{r}{r + 1}\right)^r \to \frac{1}{e} \lt 1$ : converges.
$\frac{1}{\sqrt{r}}$ decreases to 0, so Leibniz gives convergence; but $\sum\frac{1}{\sqrt{r}}$ diverges (compare with $\frac{1}{r}$ ).
$2 + 2\left(1.2 + 0.72 + \cdots\right) = 2 + 2 \times \frac{1.2}{1 - 0.6} = 2 + 6 = 8$ m.

Sources and credits

Written by Ganzaa: the explanations, the proofs, the diagrams (the partial sums are computed exactly), the worked examples, the practice items and the quiz questions.
Video: Divergence of the Harmonic Series (Exam Question 12 of 12), Eddie Woo, embedded from YouTube.

1Can infinitely many numbers add up to something finite?

2The harmonic series: terms that shrink, a sum that does not

Can infinitely many numbers add up to something finite?

The harmonic series: terms that shrink, a sum that does not