01 / 20

The \(n\)th Term Test
for Divergence

Master the quickest test in your series toolkit: if the terms don't approach zero, the series must diverge.

AP Calculus BC

What Is an Infinite Series?

An infinite series is the sum of infinitely many terms written as \(\displaystyle\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots\). We define its value through the sequence of partial sums \(S_N = \sum_{n=1}^{N} a_n\).

Before reaching for heavier tests, we need a fast first check: the \(n\)th Term Test.

Visualizing Partial Sums

N (number of terms)
\(S_N\)
\(L = 1\)
 1 2 3 4 5 6 7 8

Why Must Terms Shrink to Zero?

If a series converges, then \(S_N \to L\) and also \(S_{N-1} \to L\). Since \(a_N = S_N - S_{N-1}\), we get:

This is the key logical link: convergence of the series forces the terms to approach zero. The contrapositive gives us our test.

The \(n\)th Term Test for Divergence

Compute
\(\lim a_n\)
\(\neq 0\)?
 DIVERGES

Formal Logic Behind the Test

Direct statement
If \(\sum a_n\) converges, then \(\lim_{n\to\infty} a_n = 0\).
Contrapositive
If \(\lim_{n\to\infty} a_n \neq 0\), then \(\sum a_n\) diverges.

Necessary vs. Sufficient

\(\lim_{n\to\infty} a_n = 0\) is necessary for convergence but not sufficient. The harmonic series proves this:

\(n\)
\(1\)
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
\(\frac{1}{8}\)
\(S_N \to \infty\)
yet \(a_n \to 0\)

A One-Way Street

Series Converges IMPLIES
\(\lim_{n\to\infty} a_n = 0\)
 ✕ DOES NOT IMPLY

How to Apply the Test

1
Identify \(a_n\): Write the general term of the series explicitly.
2
Compute the limit: Evaluate \(\displaystyle\lim_{n\to\infty} a_n\) using algebra, L'Hopital's Rule, or known limits.
3
Interpret the result: If the limit is not zero (or DNE), declare divergence. If the limit equals zero, state the test is inconclusive.

Example 1: \(\sum \frac{n}{2n+1}\)

Step 1: The general term is \(a_n = \dfrac{n}{2n+1}\).

Step 2: Compute the limit:

Step 3: Since \(\lim a_n = \frac{1}{2} \neq 0\), the series diverges by the \(n\)th Term Test.

Example 2: \(\sum (-1)^n\)

Step 1: The general term is \(a_n = (-1)^n\).

Step 2: Compute the limit:

Step 3: Since the limit does not exist, the series diverges by the \(n\)th Term Test.

Example 3: \(\sum \frac{n^2}{e^n}\)

Step 1: The general term is \(a_n = \dfrac{n^2}{e^n}\).

Step 2: Compute the limit. Exponentials dominate polynomials:

Step 3: Since \(\lim a_n = 0\), the \(n\)th Term Test is inconclusive. We need a different test (the Ratio Test confirms convergence here).

Example 4: \(\sum \frac{n!}{n^n}\)

Step 1: The general term is \(a_n = \dfrac{n!}{n^n}\).

Step 2: By Stirling's approximation, \(n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\), so:

Step 3: Since \(\lim a_n = 0\), the \(n\)th Term Test is inconclusive. (The series does converge, but we'd need the Ratio Test to prove it.)

Common Mistakes to Avoid

✕ Converse Error
Claiming \(\lim a_n = 0\) proves convergence. It does not. The harmonic series is a counterexample.
✕ Forgetting DNE
If \(\lim a_n\) does not exist (e.g., oscillates), the series still diverges. Don't forget this case.
✕ Wrong Direction
Writing "converges by the \(n\)th Term Test." The test can only prove divergence, never convergence.

Summary Table

Scenario \(\lim a_n\) Conclusion
\(\sum \frac{n}{2n+1}\) \(\frac{1}{2}\) Diverges
\(\sum (-1)^n\) DNE Diverges
\(\sum \frac{1}{n}\) \(0\) Inconclusive (diverges by other test)
\(\sum \frac{1}{n^2}\) \(0\) Inconclusive (converges by other test)
\(\sum \frac{n^2}{e^n}\) \(0\) Inconclusive (converges by other test)

Question 1 of 5

What can you conclude about \(\displaystyle\sum_{n=1}^{\infty}\frac{3n+1}{n+2}\) using the \(n\)th Term Test?

Question 2 of 5

For which series does the \(n\)th Term Test prove divergence?

Question 3 of 5

A student writes: "Since \(\lim_{n\to\infty}\frac{1}{n}=0\), the harmonic series converges by the \(n\)th Term Test." What is the error?

Question 4 of 5

What is \(\displaystyle\lim_{n\to\infty}\frac{n^2+1}{3n^2-5}\), and what does the \(n\)th Term Test tell us about \(\displaystyle\sum\frac{n^2+1}{3n^2-5}\)?

Question 5 of 5

Which statement is true about the \(n\)th Term Test for Divergence?