Series and convergence

An infinite sum that has a finite answer, and the tests that decide which sums do. The BC unit that carries the most marks.

By the end of this chapter you can

Distinguish a sequence from a series
Apply the nth-term test and say what it cannot do
Recognise and sum a geometric series
Choose an appropriate convergence test

Adding infinitely many numbers should give infinity. Sometimes it does not, and that is the whole subject.

\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots = 1

$\sum_{n=1}^{\infty} 2^{- n} = 1$
$\sum_{n=0}^{\infty} 2^{- n} = 2$

checked by the build

Both are verified, and the difference between them is only where the sum starts — which is worth noticing early, because an off-by-one in the index is the most common arithmetic slip in this unit.

Sequence against series

A sequence is a list: $a_1, a_2, a_3, \ldots$
A series is the sum of a sequence: $\sum a_n$ .

They converge for different reasons, and confusing them is fatal. The sequence $\tfrac1n$ converges — to zero. The series $\sum\tfrac1n$ diverges — to infinity. Same numbers, opposite answers.

A series converges when its partial sums $S_N = \sum_{n=1}^{N} a_n$ approach a limit. The series is the limit of a sequence — the sequence of partial sums, not the sequence of terms.

The nth-term test

If $\lim_{n\to\infty} a_n \ne 0$ , the series diverges.

$\lim_{n \to \infty} \frac{1}{n} = 0$
$\lim_{n \to \infty}\left(\frac{n}{n + 1}\right) = 1$

checked by the build

$\sum \tfrac{n}{n+1}$ diverges, because its terms head for $1$ rather than $0$ — you are eventually adding roughly $1$ forever.

$\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}$

checked by the build

That is the convergent one, and its value — Euler’s famous result — is verified above. The harmonic series $\sum\tfrac1n$ has no such value, because it has no sum.

Geometric series

The one family with a closed form, and therefore the one worth knowing cold:

\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\quad\text{when } |r| < 1

and diverges when $|r| \ge 1$ .

$\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^{n} = \frac{3}{2}$
$\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^{n} = 2$
$\sum_{n=1}^{\infty} \left(\frac{1}{3}\right)^{n} = \frac{1}{2}$

checked by the build

Compare the first and third: same ratio, different starting index, and the answers differ by exactly the $n = 0$ term, which is $1$ . Starting index is not a detail.

Why $|r| < 1$ . The partial sum is $a\tfrac{1-r^N}{1-r}$ , and $r^N \to 0$ exactly when $|r| < 1$ . Outside that, $r^N$ grows or oscillates and nothing settles.

Choosing a test

The series looks like	Try
$ar^n$	geometric — and you get the exact sum
$\tfrac{1}{n^p}$	$p$ -series: converges exactly when $p > 1$
terms not going to zero	nth-term test, done
alternating signs	alternating series test
a messy rational function	limit comparison with its dominant behaviour
factorials or $n$th powers	ratio test

$\sum_{n=1}^{\infty} \frac{1}{n^{4}} = \frac{\pi^{4}}{90}$
$\sum_{n=1}^{\infty} \frac{1}{n^{3}} = \zeta\left(3\right)$

checked by the build

Both are $p$ -series with $p > 1$ , so both converge — and the contrast between them is the point. The fourth-power sum has the tidy closed form $\tfrac{\pi^4}{90}$ . The cube sum has none: the best anyone can write is $\zeta(3)$ , which is a name for the number rather than a formula for it. It is called Apéry’s constant, and proving it is even irrational took until 1978.

Writing the second claim as a decimal is what I tried first, and the checker refused it — correctly, because a truncated decimal is not equal to the sum. It would not prove the claim false either, which is exactly the unproved verdict doing its job.

So: knowing a series converges is a different question from knowing its value, and the exam almost always asks the first.