5.2 Infinite Series

Big idea

Adding infinitely many numbers is not an operation you can perform, so it has to be defined rather than carried out. The definition is the only reasonable one: add the first term, then the first two, then the first three, and watch the resulting sequence of running totals. Those totals are the partial sums, and if they converge, their limit is declared to be the sum of the series. If they do not converge, the series has no sum and is called divergent.

This is a genuine shift in what a symbol means. The expression $\sum_{n=1}^{\infty} a_n$ does not denote an addition; it denotes a limit, and it only names a number when that limit exists. Once you hold that in mind, the strange behavior of infinite sums stops being strange. An infinite sum can be finite because the partial sums are increasing and blocked by a ceiling, and an infinite sum of terms shrinking to zero can still be infinite if the shrinking is too slow.

Two families of series can be summed exactly, and both do it by making the partial sums explicit. A geometric series has a closed formula for $S_N$ that comes from a one-line algebraic trick, and taking the limit gives the sum whenever the ratio is small enough. A telescoping series has partial sums that mostly cancel, leaving only a couple of surviving terms whose limit you can read off. Almost every other series in this chapter will be decided rather than evaluated: you will learn whether it converges, without producing its value.

The first tool for deciding is the divergence test, and its logic is worth getting right from the start. If the terms do not approach zero, the running totals cannot settle, so the series diverges. The converse is false, and calling this the divergence test rather than the convergence test is a reminder of that.

Decoder

The sum of a series is the limit of its sequence of partial sums; the series converges exactly when that limit exists, and its value is that limit.

The load-bearing word is “sequence”. A series problem is a sequence problem wearing different notation, which is why the previous section was the foundation. The object you take a limit of is $S_N = a_1 + a_2 + \cdots + a_N$, a perfectly ordinary sequence indexed by how many terms you have added.

Two different sequences travel with every series and they must not be confused: the sequence of terms $\{a_n\}$ and the sequence of partial sums $\{S_N\}$. The terms of the harmonic series go to $0$ while its partial sums go to infinity, so the two can behave completely differently. Whenever a statement in this chapter sounds surprising, check which of the two sequences it is about.

The classic mistake is running the divergence test backwards. The test says: if $\lim a_n$ is not $0$, the series diverges. It never says that $\lim a_n = 0$ implies convergence, and no amount of evidence will make it say that, because the harmonic series is a permanent counterexample. When the terms go to zero the test is finished and you have learned nothing.

Definitions and results

Series and partial sums. Given a sequence $\{a_n\}$, the $N$th partial sum is $S_N = \sum_{n=1}^{N} a_n$. The series $\sum_{n=1}^{\infty} a_n$ converges to $S$ when $\lim_{N \to \infty} S_N = S$, and diverges otherwise.

Geometric series. For $a \neq 0$, the series $\sum_{n=0}^{\infty} a r^{n}$ converges if and only if $|r| < 1$, and then

$$ \sum_{n=0}^{\infty} a r^{n} = \frac{a}{1 - r} $$

In words: first term divided by one minus the ratio. For $|r| \ge 1$ the terms do not shrink to zero and the series diverges.

Where the formula comes from. Write $S_N = a + ar + \cdots + ar^{N-1}$ and subtract $rS_N$. Everything cancels except the ends, giving $S_N(1 - r) = a\left(1 - r^{N}\right)$, so $S_N = \frac{a\left(1 - r^N\right)}{1 - r}$. When $|r| < 1$ the term $r^N$ dies and the limit is $\frac{a}{1-r}$.

Index care. The formula needs the actual first term in the numerator, whatever the starting index is. Writing out the first two terms and dividing to find $r$ is safer than reading exponents.

Telescoping series. If $a_n = b_n - b_{n+k}$ for a fixed $k$, then almost every term of $S_N$ cancels against another. What remains is a fixed block from the front and a block from the back, and the series converges exactly when $\{b_n\}$ converges.

The divergence test. If $\lim_{n \to \infty} a_n$ fails to exist or is not $0$, then $\sum a_n$ diverges. Equivalently, if $\sum a_n$ converges then $a_n \to 0$.

What the divergence test cannot do. It can never establish convergence. The harmonic series $\sum \frac1n$ and the series $\sum \frac{1}{n^2}$ both have terms tending to $0$; the first diverges and the second converges. Any test that looks only at $\lim a_n$ must be silent about both.

Algebra of series. If $\sum a_n$ and $\sum b_n$ converge, then $\sum (a_n \pm b_n)$ converges to the sum or difference of the values, and $\sum c\,a_n$ converges to $c$ times the value. Adding, removing, or changing finitely many terms changes the value but never the convergence.

Worked examples

A geometric series with a shifted index

Evaluate $\displaystyle\sum_{n=1}^{\infty} \frac{2^{\,n+1}}{5^{\,n}}$.

Do not guess the first term from the exponents; compute it. At $n = 1$ the term is $\frac{4}{5}$, and at $n = 2$ it is $\frac{8}{25}$. The ratio is

$$ \frac{8/25}{4/5} = \frac{8}{25}\cdot\frac{5}{4} = \frac{2}{5} $$

so the series is geometric with $a = \frac45$ and $r = \frac25$. Since $|r| < 1$ it converges, and

$$ \sum_{n=1}^{\infty}\frac{2^{\,n+1}}{5^{\,n}} = \frac{4/5}{1 - 2/5} = \frac{4/5}{3/5} = \frac{4}{3} $$

Check the plausibility: the first three terms are $0.8 + 0.32 + 0.128 = 1.248$, already close to $1.333$ with the remaining tail small.

A repeating decimal as a series

Write $0.\overline{27}$ as a fraction.

The decimal means $0.27 + 0.0027 + 0.000027 + \cdots$, which is geometric with $a = \frac{27}{100}$ and $r = \frac{1}{100}$. Therefore

$$ 0.\overline{27} = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11} $$

Dividing $3$ by $11$ returns $0.272727\ldots$, so the answer checks. Every repeating decimal is a geometric series, which is why every one of them is rational.

A telescoping sum done through the partial sums

Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+2)}$.

Partial fractions give $\frac{1}{n(n+2)} = \frac12\left(\frac1n - \frac{1}{n+2}\right)$, which you can confirm by recombining: $\frac12\cdot\frac{(n+2) - n}{n(n+2)} = \frac{1}{n(n+2)}$.

Now write the partial sum and look for the cancellation. Each $-\frac{1}{n+2}$ cancels the $\frac{1}{m}$ two places later, so only the first two positive terms and the last two negative terms survive:

$$ S_N = \frac12\left(1 + \frac12 - \frac{1}{N+1} - \frac{1}{N+2}\right) $$

As $N \to \infty$ the last two pieces vanish, so the series converges and

$$ \sum_{n=1}^{\infty}\frac{1}{n(n+2)} = \frac12\cdot\frac32 = \frac{3}{4} $$

Check numerically: the first four terms are $\frac13 + \frac18 + \frac{1}{15} + \frac{1}{24} \approx 0.3333 + 0.1250 + 0.0667 + 0.0417 = 0.5667$, and the formula gives $S_4 = \frac12\left(1.5 - \frac15 - \frac16\right) = \frac12(1.1333) = 0.5667$. The two agree.

Using and failing to use the divergence test

Consider $\displaystyle\sum_{n=1}^{\infty}\frac{4n^2 + 1}{3n^2 - n}$ and $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}$.

For the first, divide through by $n^2$: the terms approach $\frac43$, which is not $0$. The partial sums therefore grow by about $\frac43$ every step and cannot settle, so the series diverges. That took one limit and no further work.

For the second, the terms approach $0$, so the test returns nothing at all. This is not a hint of convergence; it is silence. The harmonic series in fact diverges, and grouping shows why: $\frac13 + \frac14 > \frac12$, then $\frac15 + \cdots + \frac18 > \frac12$, then the next eight terms exceed $\frac12$, and so on forever. The partial sums pass every ceiling, slowly but without stopping. Deciding cases like this one is what the next four sections are for.

Practice

Start with the family you can sum exactly. Identify the first term and the ratio, check that the ratio is smaller than $1$ in size, and apply the formula, including on repeating decimals and shifted indices.

Then the first decision tool. Take the limit of the terms, conclude divergence when it is not zero, and say explicitly that the test is inconclusive when it is.

Quiz

Five items on partial sums, geometric and telescoping series, and the correct use of the divergence test.