Calculus 2Sequences and Series

The Divergence Test

If the terms do not go to 0, the series diverges.

Not started Jump to practice

Before any hard work on a series, check one thing: do its terms go to 0? If not, you are done. This test follows straight from the definition of partial sums.

Why it works

Each term is the difference of two neighboring partial sums: $a_n = S_n - S_{n-1}$. If the series converges to $S$, both $S_n$ and $S_{n-1}$ approach $S$, so $a_n \to S - S = 0$. Turned around: if the terms do not go to 0, the series cannot converge.

Key idea

Divergence test: if $\lim_{n\to\infty} a_n \ne 0$ (or the limit does not exist), then $\sum a_n$ diverges. If $\lim a_n = 0$, the test says nothing.

Worked example: Terms with a nonzero limit

Does $\sum_{n=1}^{\infty}\frac{n}{2n + 1}$ converge?
  1. $\frac{n}{2n + 1} \to \frac{1}{2}$, not 0.
  2. So the series diverges: each new term adds about $\frac{1}{2}$, and $S_N$ grows roughly like $\frac{N}{2}$.

Answer: It diverges.

For $\sum\frac{n}{2n + 1}$ the bars (terms) level off near $\frac{1}{2}$ instead of shrinking, so the partial sums climb in nearly equal steps forever.

Worked example: A limit you have to work for

Does $\sum_{n=1}^{\infty}\cos\frac{1}{n}$ converge?
  1. $\frac{1}{n} \to 0$ and cosine is continuous, so $\cos\frac{1}{n} \to \cos 0 = 1$.
  2. The terms do not go to 0, so the series diverges.

Answer: It diverges.

Common mistake

Using the test to claim convergence. "The terms of $\sum\frac{1}{n}$ go to 0, so it converges" is wrong: the harmonic series diverges. And $\sum\frac{1}{n^{2}}$, whose terms also go to 0, converges. When $a_n \to 0$ the test is silent; move on to another test. The test can only ever say "diverges".

Worked example: When the test is silent

What does the divergence test say about $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$?
  1. $\frac{1}{\sqrt{n}} \to 0$.
  2. So the test gives no conclusion. (The The Integral Test and p-Series will show this series diverges.)

Answer: Nothing.

Worked example: A limit that is a famous number

Does $\sum_{n=1}^{\infty}\left(1 + \frac{1}{n}\right)^{n}$ converge?
  1. The terms approach $e \approx 2.718$, the limit that defines $e$.
  2. So each term adds almost $2.7$ to the running total, and the series diverges.

Answer: It diverges.

Make the divergence test your first move on every series: it takes one limit. If the terms do not go to 0, stop. If they do, you have learned that you need a sharper tool, such as the integral, comparison or ratio test, which come next.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 1 XP.

Loading practice…

Builds on

Leads to

Sources and further reading

  • OpenStax Calculus Volume 2, Section 5.3: The Divergence and Integral Tests
  • Paul’s Online Math Notes (Lamar University), Calculus II: Convergence/Divergence of Series (divergence test)
@teddyrookbook GitHub