The nth Term Test for Divergence — AP® Calculus BC Notes

Every later test in Unit 10 takes some work. The nth term test takes almost none, so it is the first thing to check. It looks only at the terms $a_n$ and asks one question: do they approach 0?

The test

nth term test for divergence. If $\lim_{n \to \infty} a_n \ne 0$ , or the limit does not exist, then $\sum a_n$ diverges.

Why it works: if a series converges to $S$ , then its partial sums $S_n$ and $S_{n-1}$ both approach $S$ , so

$a_n = S_n - S_{n-1} \to S - S = 0.$

So every convergent series has terms that approach 0. A series whose terms do not approach 0 cannot converge.

Two plots for n = 1 to 15, with blue bars for the terms and vermillion dots for the partial sums. Left: a sub n = n/(2n + 1); the bars level off near 1/2, not 0, and the partial sums climb by about 1/2 each step, past 7 by n = 15: the series diverges by the nth term test. Right: a sub n = 1/n; the bars shrink toward 0, yet the partial sums keep growing, past 3.3 by n = 15; terms approaching 0 do not settle the question, so the nth term test is silent. — **Figure 1.** Left: terms approach $\frac{1}{2}$ . Right: terms approach 0. LevlPrep original diagram.

The left panel shows why: when the terms level off at $\frac{1}{2}$ , each new term adds about $\frac{1}{2}$ , and the partial sums climb without bound.

What it cannot do

The test only ever proves divergence. If $\lim a_n = 0$ , it is inconclusive: the series might converge or diverge. The right panel shows terms $\frac{1}{n}$ that approach 0 while the partial sums keep growing. In 10.1 you saw another: $\sum \ln\left(\frac{n + 1}{n}\right)$ has terms approaching 0 and partial sums $\ln(n + 1)$ that grow without bound. On the other hand $\sum \left(\frac{1}{2}\right)^n$ also has terms approaching 0, and it converges.

$\lim_{n \to \infty} a_n$	What the nth term test says
a number other than 0	the series diverges
does not exist (including $\pm\infty$ )	the series diverges
$0$	nothing: use another test

Using it

Worked example 1. Does $\sum_{n=1}^{\infty} \frac{3n^2 - 1}{5n^2 + n}$ converge?

Limit of the terms: divide the top and bottom by $n^2$ : $\frac{3 - 1/n^2}{5 + 1/n} \to \frac{3}{5}$ .

Conclusion: $\lim_{n \to \infty} \frac{3n^2 - 1}{5n^2 + n}$ $= \frac{3}{5} \ne 0$ , so the series diverges by the nth term test.

Worked example 2. What does the nth term test say about $\sum_{n=1}^{\infty} (-1)^n \frac{2n}{n + 3}$ and $\sum_{n=1}^{\infty} \frac{\ln n}{n}$ ?

First: the sizes $\frac{2n}{n + 3}$ approach 2, so the terms jump between values near 2 and near $-2$ . The limit does not exist, so it is not 0: the series diverges by the nth term test.

Second: $\frac{\ln x}{x}$ is an $\frac{\infty}{\infty}$ form, and L'Hospital's rule gives $\frac{1/x}{1} \to 0$ . The terms approach 0, so the nth term test says nothing. (Topic 10.4 settles it.)

Saying it right

A justification names the limit, compares it with 0, names the test and concludes about the series:

" $\lim_{n \to \infty} \frac{3n^2 - 1}{5n^2 + n}$ $= \frac{3}{5} \ne 0$ , so $\sum \frac{3n^2 - 1}{5n^2 + n}$ diverges by the nth term test."

Three phrases cost points: "the terms diverge" when they approach $\frac{3}{5}$ ; "the series converges to $\frac{3}{5}$ ", which mixes up the terms with the sum; and "the terms go to 0, so the series converges", which uses the test backward.

Limits in this unit are in $n$ : write $\lim_{n \to \infty} a_n$ . To use L'Hospital's rule or a derivative, switch to a function $f(x)$ with $f(n) = a_n$ , find $\lim_{x \to \infty} f(x)$ , and then state the result for $n$ .