1.5.3 Infinite series and when they converge

What an infinite sum means

A series with no last term, such as $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ , cannot be added up term by term. Instead, add the first $n$ terms to get the partial sum $S_n$ (lesson 1.5.2). Then ask whether the sequence $S_1, S_2, S_3, \ldots$ has a limit (lesson 1.5.1).

$n$	1	2	3	5	10	20
$S_n$	0.5	0.75	0.875	0.96875	0.99902	0.99999905
$1 - S_n$	0.5	0.25	0.125	0.03125	0.00098	0.00000095

The partial sums close in on 1, and the gap $1 - S_n$ halves with every term: it is $\frac{1}{2^n}$ . A square shows why.

Each piece fills half of what is still empty. After $n$ pieces the empty part is $\frac{1}{2^n}$ , so the filled area closes in on the whole square, 1.

Rule: Sum of an infinite series

\sum_{k=1}^{\infty} a_k = \lim_{n \to \infty} S_n

If the partial sums have a limit, the series converges and that limit is its sum. If they have none, the series diverges and has no sum. The sum has the same units as the terms.

1 - S_n

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The infinite geometric series

Lesson 1.5.2 gave the partial sums of a geometric series with first term $a$ and ratio $r$ :

S_n = a\,\frac{1 - r^n}{1 - r}

Only $r^n$ changes as $n$ grows. When $|r| < 1$ , lesson 1.5.1 showed that $r^n \to 0$ , so $S_n$ settles on $\frac{a}{1 - r}$ . The gap left after $n$ terms is the part that vanishes: $\frac{a\,r^n}{1 - r}$ .

Rule: Infinite geometric series

\begin{gathered} a + ar + ar^2 + \cdots = \frac{a}{1 - r} \\ \text{only when } |r| < 1 \end{gathered}

The first term over one minus the ratio. After $n$ terms the partial sum still misses by $\frac{a\,r^n}{1 - r}$ , which shrinks to 0.

For $\frac{1}{2} + \frac{1}{4} + \cdots$ , $a = \frac{1}{2}$ and $r = \frac{1}{2}$ , so the sum is $\frac{1/2}{1 - 1/2} = 1$ . The gap after $n$ terms is $\frac{(1/2)(1/2)^n}{1/2} = \frac{1}{2^n}$ , as in the table.

A table and the algebra

Take $3 + \frac{3}{4} + \frac{3}{16} + \cdots$ , with $a = 3$ and $r = \frac{1}{4}$ . The partial sums are 3, 3.75, 3.9375, 3.984375, 3.99609375, and after 10 terms 3.99999619. The algebra gives the limit: $\frac{3}{1 - 1/4} = 4$ .

Repeating decimals

A repeating decimal is an infinite geometric series. $0.999\ldots$ is $0.9 + 0.09 + 0.009 + \cdots$ , with $a = 0.9$ and $r = 0.1$ , so it equals $\frac{0.9}{1 - 0.1} = 1$ exactly.

$0.272727\ldots$ is $0.27 + 0.0027 + \cdots$ , with $a = 0.27$ and $r = 0.01$ . Its sum is $\frac{0.27}{0.99} = \frac{27}{99} = \frac{3}{11}$ .

3 + \frac{3}{4} + \cdots

import numpy as np def infinite_geometric(a, r): return a / (1 - r) partial = np.cumsum(3 * 0.25 ** np.arange(10)) print(partial[:6]) print(partial[9], infinite_geometric(3, 0.25)) print(infinite_geometric(0.9, 0.1)) print(infinite_geometric(0.27, 0.01), 3 / 11)

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Every ratio

A negative ratio gives an alternating series. $1 - \frac{1}{2} + \frac{1}{4} - \cdots$ has $a = 1$ and $r = -\frac{1}{2}$ , so its sum is $\frac{1}{1 - (-1/2)} = \frac{2}{3}$ . Its partial sums 1, 0.5, 0.75, 0.625, 0.6875 land above and below $\frac{2}{3}$ in turn, ever closer.

Geometric partial sums settle on $\frac{a}{1 - r}$ , from one side or from both. The harmonic sums keep rising by about 2.3 every time $n$ is multiplied by 10, while the sums of $\frac{1}{n^2}$ level off.

When $|r| \ge 1$ the series diverges. The terms never shrink, so the partial sums never settle.

$r$	series	partial sums	what they do
1	$1 + 1 + 1 + \cdots$	1, 2, 3, 4, 5	grow without end
$-1$	$1 - 1 + 1 - \cdots$	1, 0, 1, 0, 1	bounce for ever
2	$1 + 2 + 4 + \cdots$	1, 3, 7, 15, 31	grow without end
$-2$	$1 - 2 + 4 - \cdots$	1, $-1$ , 3, $-5$ , 11	swing wider and wider

The formula gives an answer anyway, and it is wrong. For $r = 2$ it says $\frac{1}{1 - 2} = -1$ , a negative sum of positive terms. For $r = -1$ it says $\frac{1}{2}$ , a value the partial sums never take.

Rule: Terms must shrink to 0

\begin{gathered} a_n \not\to 0 \;\Longrightarrow\; \text{the series diverges} \end{gathered}

If the terms do not tend to 0, each new term keeps moving the partial sum by a sizeable amount, so it cannot settle. Check this before reaching for a formula.

Try it The cell prints the first six partial sums for each ratio in the text: one that converges and the four that do not. NumPy prints the whole-number cases without decimal points.

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Check yourself

Which of these series has a sum?

1 + 1.1 + 1.21 + 1.331 + \cdots

1 - 1 + 1 - 1 + \cdots

5 - 4.5 + 4.05 - \cdots

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots

Try every ratio yourself. Slide $r$ from 0.9 up to 1 and past it, then to $-1$ and below, and watch the dots settle, climb or bounce. The harmonic series, the subject of the next section, is one of the choices too.

Explore

Each dot is a partial sum: the total of the first n terms. Change the series and the ratio and see which totals settle on a number.

Ratio r = 0.5

First term a = 1

Terms added n = 10

Term number 10: 0.001953
Total of 10 terms: S10 = 1.998047
Still to go: 0.001953
Verdict: The partial sums settle on 2.

Shrinking terms are not enough

The harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ has terms that shrink to 0. Yet its partial sums grow without end, only very slowly.

$n$	10	100	1,000	10,000	100,000	1,000,000
harmonic $S_n$	2.929	5.187	7.485	9.788	12.090	14.393
$1 + \frac{1}{4} + \frac{1}{9} + \cdots$	1.549768	1.634984	1.643935	1.644834	1.644924	1.644933

Each tenfold increase in $n$ adds about 2.30 to the harmonic sum, close to $\ln 10$ (lesson 1.4.3). The sum passes any number you name, given enough terms, so the series diverges. The Proof Gym (stage PG5) proves it.

Squaring the bottoms changes everything. $1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \cdots$ converges: its partial sums level off at $\frac{\pi^2}{6} \approx 1.644934$ . Both series have terms that shrink to 0; only one has a sum.

Rule: Shrinking terms are necessary, not enough

\begin{gathered} a_n \to 0 \;\Longrightarrow\; \text{check further} \end{gathered}

Terms that tend to 0 allow a sum but do not promise one. Compare with a series you know, or watch the partial sums as $n$ is multiplied by 10.

n^2

import math import numpy as np n = np.arange(1, 1_000_001, dtype=float) harmonic = np.cumsum(1 / n) squares = np.cumsum(1 / n**2) for k in [10, 100, 1_000, 10_000, 100_000, 1_000_000]: print(k, round(harmonic[k - 1], 3), round(squares[k - 1], 6)) print(round(math.pi**2 / 6, 6))

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How many terms for a given accuracy

In practice you stop after $n$ terms and accept a small error. For a geometric series that error is the gap from the rule above. Logarithms (lesson 1.4.4) then say how many terms make it smaller than a tolerance.

Example 1. For $\frac{1}{2} + \frac{1}{4} + \cdots$ , the gap after $n$ terms is $0.5^n$ . To be within $10^{-6}$ of 1, solve $0.5^n < 10^{-6}$ . Taking logs gives $n \ln 0.5 < \ln 10^{-6}$ .

Now divide by $\ln 0.5$ , which is negative, so the inequality flips (lesson 0.3.3): $n > \frac{\ln 10^{-6}}{\ln 0.5} \approx 19.93$ . The first whole number is 20. Check: the gap is $1.9 \times 10^{-6}$ after 19 terms and $9.5 \times 10^{-7}$ after 20.

Example 2. For $1 + 0.9 + 0.81 + \cdots$ , the sum is $\frac{1}{1 - 0.9} = 10$ and the gap is $\frac{0.9^n}{0.1} = 10 \times 0.9^n$ . Within 0.01 needs $0.9^n < 0.001$ , so $n > \frac{\ln 0.001}{\ln 0.9} \approx 65.56$ : 66 terms. A ratio close to 1 converges slowly.

Rule: Terms needed for a tolerance

\begin{gathered} \text{gap after } n \text{ terms} = \frac{|a|\,|r|^n}{1 - r} \\ \text{gap} < \text{tol} \ \text{ when} \\ n > \frac{\ln\big(\text{tol}\,(1 - r) / |a|\big)}{\ln |r|} \end{gathered}

For $|r| < 1$ . Take the first whole number above the bound. A negative ratio uses its size $|r|$ ; the bottom $1 - r$ stays as it is.

Without a formula for the gap, watch it numerically. For $1 + \frac{1}{4} + \frac{1}{9} + \cdots$ the gap after $n$ terms is close to $\frac{1}{n}$ : 0.0952 after 10 terms, 0.00995 after 100, 0.0009995 after 1,000. Three correct decimals cost a thousand terms.

1 - \frac{1}{2} + \cdots

import math import numpy as np def terms_needed(a, r, tol): n = 0 while abs(a * r**n / (1 - r)) >= tol: n = n + 1 return n print(math.log(1e-6) / math.log(0.5), terms_needed(0.5, 0.5, 1e-6)) print(math.log(0.001) / math.log(0.9), terms_needed(1, 0.9, 0.01)) print(terms_needed(1, -0.5, 0.001)) k = np.arange(1, 1001, dtype=float) squares = np.cumsum(1 / k**2) for n in [10, 100, 1000]: print(n, round(math.pi**2 / 6 - squares[n - 1], 7))

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Case study 1.5.5 uses the infinite geometric series to price a payment that never stops. Level 2 writes functions such as $e^x$ as infinite series (module 2.7), and the question of how many terms you need returns there.

Common mistakes

$\frac{a}{1 - r}$ needs $|r| < 1$ . $1 + 2 + 4 + \cdots$ has no sum, even though $\frac{1}{1 - 2} = -1$ .
$a$ is the first term actually added. For $\frac{1}{2} + \frac{1}{4} + \cdots$ , $a = \frac{1}{2}$ , not 1, and the sum is 1, not 2.
Keep the sign of a negative ratio. $1 - \frac{1}{2} + \frac{1}{4} - \cdots$ is $\frac{1}{1 + 1/2} = \frac{2}{3}$ , not $\frac{1}{1 - 1/2} = 2$ .
Shrinking terms do not guarantee a sum, as the harmonic series shows. They are needed, but not enough.
A partial sum is not the sum. $S_{20}$ of $\frac{1}{2} + \frac{1}{4} + \cdots$ is 0.99999905, close to 1 but not 1.
Flip the inequality when you divide by $\ln |r|$ . For $|r| < 1$ it is negative, so $<$ becomes $>$ .

Infinite series

0 of 12 correct

1.
Find the partial sum $S_3$ of $\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8} + \cdots$
2.
Find $\displaystyle\sum_{k=0}^{\infty} 2\left(\tfrac{1}{2}\right)^k$
3.
Find $10 + 2 + 0.4 + \cdots$
4.
Find $\displaystyle\sum_{k=1}^{\infty} \left(\tfrac{1}{3}\right)^k$
5.
Find $6 - 2 + \tfrac{2}{3} - \cdots$
6.
Write $0.272727\ldots$ as a fraction
7.
An infinite geometric series has first term 6 and sum 8. Find its ratio $r$
8.
For which ratio $r$ does $1 + r + r^2 + \cdots$ add to 4?
9.
Find the partial sum $S_{101}$ of $1 - 1 + 1 - 1 + \cdots$
10.
Find the partial sum $S_4$ of the harmonic series $1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots$
11.
How far is $S_{10}$ of $\tfrac{1}{2} + \tfrac{1}{4} + \cdots$ from the sum 1?
12.
How many terms of $1 + 0.9 + 0.81 + \cdots$ bring the partial sum within 0.01 of its sum, 10?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

S_n

Infinite series and when they converge

What an infinite sum means

Rule: Sum of an infinite series

The infinite geometric series

Rule: Infinite geometric series

A table and the algebra

Repeating decimals

Every ratio

Rule: Terms must shrink to 0

Shrinking terms are not enough

Rule: Shrinking terms are necessary, not enough

How many terms for a given accuracy

Rule: Terms needed for a tolerance

Common mistakes

Infinite series

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