Convergent and Divergent Series — AP® Calculus BC Notes

An infinite series is a sum with no last term:

$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$

You cannot add infinitely many numbers one at a time, so "the sum" needs a definition. It comes from adding a finite number of terms and asking where those totals head.

Partial sums

The $n$ th partial sum $S_n$ is the sum of the first $n$ terms:

$S_1 = a_1, \quad S_2 = a_1 + a_2, \quad S_3 = a_1 + a_2 + a_3, \quad S_n = \sum_{k=1}^{n} a_k.$

The partial sums form a new sequence. The series converges to the number $S$ if $\lim_{n \to \infty} S_n = S$ ; then $S$ is the sum of the series. If the partial sums have no finite limit, because they grow without bound or never settle, the series diverges.

For the series with terms a sub n = 1 over n(n + 1), n = 1 to 12: blue bars show each term, starting at 1/2, then 1/6, 1/12 and shrinking toward 0; emerald dots show each partial sum S sub n = n/(n + 1): 1/2, 2/3, 3/4 and so on, rising toward the dashed line at height 1. The sequence of partial sums has limit 1, so the series converges and its sum is 1. — **Figure 1.** Terms (bars) and partial sums (dots) of $\sum \frac{1}{n(n + 1)}$ . LevlPrep original diagram.

Two sequences, two questions. In the figure, the terms $a_n$ shrink toward 0 and the partial sums $S_n$ climb toward 1. The series converges, and its sum is 1. Say it that way: "the series converges to 1". The 2024 exam readers took points from answers that said the terms converge when they meant the series.

Finding a sum from the partial sums

Worked example 1. Find $\sum_{n=1}^{\infty} \frac{1}{n(n + 1)}$ .

Split each term: $\frac{1}{n(n + 1)}$ $= \frac{1}{n} - \frac{1}{n + 1}$ (check by combining the right side).

Write out a partial sum: $S_n$ $= \left(1 - \frac{1}{2}\right)$ $+ \left(\frac{1}{2} - \frac{1}{3}\right)$ $+ \cdots$ $+ \left(\frac{1}{n} - \frac{1}{n + 1}\right)$ .

Each negative piece cancels the next positive piece, so $S_n$ $= 1 - \frac{1}{n + 1}$ $= \frac{n}{n + 1}$ .

Take the limit: $\lim_{n \to \infty} \frac{n}{n + 1}$ $= 1$ . The series converges, and its sum is 1.

A series like this, where the partial sums collapse, is called telescoping. Most series do not have a neat formula for $S_n$ ; the tests in 10.3 through 10.9 decide convergence without one.

Going the other way, if you know the partial sums you know the terms: $a_1 = S_1$ and $a_n = S_n - S_{n-1}$ . If $S_n = \frac{3n}{n + 2}$ , then $a_3$ $= S_3 - S_2$ $= \frac{9}{5} - \frac{3}{2}$ $= \frac{3}{10}$ , and the sum is $\lim S_n = 3$ .

Three ways to diverge

Series	Partial sums	Why it diverges
$1 + 1 + 1 + \cdots$	$S_n = n$	grow without bound
$1 - 1 + 1 - 1 + \cdots$	$1, 0, 1, 0, \ldots$	never settle on one number
$\sum \ln\left(\frac{n + 1}{n}\right)$	$S_n = \ln(n + 1)$	grow without bound, slowly, though the terms approach 0

The middle row is a warning about regrouping. Grouping as $(1 - 1) + (1 - 1) + \cdots$ suggests 0; grouping as $1 + (-1 + 1) + \cdots$ suggests 1. Each grouping looks at only some of the partial sums. The definition looks at all of them, and they have no limit, so the series has no sum.

The last row is the most important fact in this unit: terms that approach 0 do not guarantee convergence. Infinitely many small pieces can still add up to something infinite.

Rules that carry over

Worked example 2. $\sum_{n=1}^{\infty} a_n$ $= 4$ and $a_1 + a_2 + a_3 = 2.5$ . Find $\sum_{n=1}^{\infty} 3a_n$ and $\sum_{n=4}^{\infty} a_n$ .

Multiplying every term by 3 multiplies every partial sum by 3, so $\sum 3a_n = 3 \cdot 4 = 12$ .

Dropping the first three terms subtracts 2.5 from every partial sum after the third, so $\sum_{n=4}^{\infty} a_n$ $= 4 - 2.5$ $= 1.5$ .

If $\sum a_n = A$ and $\sum b_n = B$ , then $\sum (a_n + b_n) = A + B$ and $\sum c\,a_n = cA$ .
If $\sum a_n$ converges and $\sum b_n$ diverges, $\sum (a_n + b_n)$ diverges.
Adding or removing finitely many terms changes the sum, never whether the series converges. That is why the tests later in the unit only care about the long-run behavior of the terms.