What could it mean to add infinitely many numbers? You cannot finish the addition. Instead, look at the running totals. This turns an infinite sum into a sequence whose limit you can study, written with sigma notation.
Why it works
For a series $\sum_{n=1}^{\infty} a_n$, the partial sums are $S_1 = a_1$, $S_2 = a_1 + a_2$, and in general $S_N = a_1 + \cdots + a_N$. These are ordinary finite sums. We define the sum of the series to be $\lim_{N\to\infty} S_N$ when that limit exists. So "the series converges" means "the running totals settle on a number", and that number is what the infinite sum means.
Key idea
$$\sum_{n=1}^{\infty} a_n = \lim_{N\to\infty} S_N, \qquad S_N = \sum_{n=1}^{N} a_n.$$
The series converges if the limit exists and diverges if it does not. Terms going to 0 is necessary for convergence, but not enough.
Worked example: Partial sums with a pattern
Find the sum of $\sum_{n=1}^{\infty}\frac{1}{n(n + 1)}$.
- The first partial sums: $S_1 = \frac{1}{2}$, $S_2 = \frac{1}{2} + \frac{1}{6} = \frac{2}{3}$, $S_3 = \frac{2}{3} + \frac{1}{12} = \frac{3}{4}$.
- The pattern is $S_N = \frac{N}{N + 1} = 1 - \frac{1}{N + 1}$ (because $\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$, and the terms cancel in pairs).
- $S_N \to 1$, so the series converges to 1.
Answer: $1$
Worked example: The harmonic series diverges
Show that $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges.
- Group the terms: $$\begin{aligned} &1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \cdots + \frac{1}{8}\right) \\ &\quad + \left(\frac{1}{9} + \cdots + \frac{1}{16}\right) + \cdots \end{aligned}$$
- Each group is at least $\frac{1}{2}$: for example $\frac{1}{5} + \cdots + \frac{1}{8} \ge 4\cdot\frac{1}{8} = \frac{1}{2}$.
- So $S_{2^{k}} \ge 1 + \frac{k}{2}$, which grows without bound. The partial sums do not settle.
Answer: It diverges.
Common mistake
Deciding that a series converges because its terms go to 0. The harmonic series has $\frac{1}{n} \to 0$ and still diverges. Terms shrinking only means each new step is small; many small steps can still add up to any amount. Check the partial sums (or use a test), not just the terms.
Worked example: Partial sums that bounce
Does $\sum_{n=0}^{\infty}(-1)^{n} = 1 - 1 + 1 - 1 + \cdots$ converge?
- $S_1 = 1$, $S_2 = 0$, $S_3 = 1$, $S_4 = 0$, and so on.
- The partial sums never settle, so the series diverges. (Grouping the terms in pairs to get 0 changes the series; the definition uses partial sums.)
Answer: It diverges.
Two facts you will use often: multiplying every term by a constant multiplies the sum; and changing or dropping finitely many terms never changes whether a series converges (only, possibly, its sum).