Sequences, partial sums, and what a series means

Understand the idea

A sequence is a list of values indexed by integers. An infinite series is studied through a different sequence: its partial sums. The series converges when those partial sums approach a finite number. This distinction separates the behavior of individual terms from the behavior of their accumulated total.

For terms $\displaystyle a_n$ , the partial sum $\displaystyle S_N$ adds the first $\displaystyle N$ terms. A convergent series must have $\displaystyle a_n\to0$ , because consecutive partial sums eventually differ by very little. The converse is false: terms can approach zero while their cumulative sum grows without bound.

Telescoping expressions illustrate the definition directly. Expand several terms to identify cancellation, write a formula for the finite partial sum, and then take its limit. Keep the initial index: changing where a series starts changes its sum by finitely many terms, though not whether it converges.

S_N=\sum_{n=1}^{N}a_n,\qquad \sum_{n=1}^{\infty}a_n=S\ \text{means}\ \lim_{N\to\infty}S_N=S

A method you can apply

Distinguish the term formula $\displaystyle a_n$ from the partial sum $\displaystyle S_N$ .
Look for a finite-sum identity, cancellation, or another suitable convergence argument.
Take the limit of the partial sums and state whether it is finite.

Worked example

Find the sum of $\displaystyle \sum_{n=1}^{\infty}1/[n(n+1)]$ .

Rewrite $\displaystyle 1/[n(n+1)]=1/n-1/(n+1)$ .
The finite sum is $\displaystyle (1-1/2)+(1/2-1/3)+\cdots+(1/N-1/(N+1))$ .
After cancellation, $\displaystyle S_N=1-1/(N+1)$ .

The partial sums approach

\displaystyle 1

, so the series converges to

\displaystyle 1

.

Try it yourself

Find $\displaystyle \sum_{n=1}^{\infty}\left(1/(n+1)-1/(n+2)\right)$ .

Show solution and reasoning

The partial sum is $\displaystyle 1/2-1/(N+2)$ , which approaches $\displaystyle 1/2$ . The starting term matters: the sum is not $\displaystyle 1$ .