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Amy Ferguson Moncure

Calculus BC · Unit 10

The nth-Term Test for Divergence

If the terms do not go to 0, the series diverges. If they do go to 0, you still need another test.

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Key Ideas Easy Medium Hard

\(\lim a_n\neq 0 \Rightarrow \sum a_n\ \text{diverges}\)

What to Remember

A nonzero limit means divergence

If \(\lim a_n\) is not 0, then \(\sum a_n\) diverges.

A zero limit decides nothing

If \(\lim a_n=0\), the nth-term test is inconclusive. The series might converge or diverge.

The test never proves convergence

Passing the “terms go to 0” check is required for convergence, but it is not enough.

Rationalize before you take the limit

A difference such as \(\sqrt{n^2+n}-n\) can look like it goes to 0. Multiply by the conjugate before you decide.

Level 1

Easy

A limit you can read from the leading terms.

Example
Does \(\sum_{n=1}^{\infty}\dfrac{n}{n+1}\) converge or diverge?

\(\lim_{n\to\infty}\dfrac{n}{n+1}=1\)

The limit is not 0.

The series diverges.

Answer: diverges

1
Does \(\sum_{n=1}^{\infty}\dfrac{n^2}{2n^2+1}\) converge or diverge?
Show Answer

\(\lim_{n\to\infty}\dfrac{n^2}{2n^2+1}=\dfrac{1}{2}\)

The limit is not 0.

The series diverges.

Answer: diverges

2
What does the nth-term test say about \(\sum_{n=0}^{\infty}\left(\dfrac{1}{2}\right)^n\)?
Show Answer

\(\lim_{n\to\infty}\left(\dfrac{1}{2}\right)^n=0\)

The test is inconclusive.

A different test is needed. This particular series is geometric and converges, but the nth-term test does not say so.

Answer: inconclusive

3
What does the nth-term test say about \(\sum_{n=1}^{\infty}\dfrac{1}{n}\)?
Show Answer

\(\lim_{n\to\infty}\dfrac{1}{n}=0\)

The test is inconclusive.

The harmonic series actually diverges, which shows why a zero limit is not enough.

Answer: inconclusive

4
Does \(\sum_{n=0}^{\infty}(-1)^n\) converge or diverge?
Show Answer

The terms are \(1,-1,1,-1,\ldots\).

The limit does not exist, so it is not 0.

The series diverges.

Answer: diverges

5
Which series must diverge by the nth-term test?
A. \(\sum 1/n^2\)
B. \(\sum n/(2n+3)\)
C. \(\sum 1/2^n\)
D. \(\sum (-1)^n/n\)
Show Answer

A, C, and D all have terms going to 0, so the test is inconclusive for them.

\(\lim \dfrac{n}{2n+3}=\dfrac{1}{2}\neq 0\)

B diverges by the nth-term test.

Answer: B. \(\sum n/(2n+3)\)

Level 2

Medium

A limit that needs algebra, or a famous sequence that does not go to 0.

Example
Does \(\sum_{n=1}^{\infty}\left(\sqrt{n^2+n}-n\right)\) converge or diverge?

Multiply by the conjugate.

\(\sqrt{n^2+n}-n=\dfrac{n}{\sqrt{n^2+n}+n}\)

\(=\dfrac{1}{\sqrt{1+1/n}+1}\)

The limit is \(\dfrac{1}{2}\).

The limit is not 0, so the series diverges.

Answer: diverges

1
Does \(\sum_{n=1}^{\infty}\left(1+\dfrac{1}{n}\right)^n\) converge or diverge?
Show Answer

\(\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e\)

\(e\neq 0\)

The series diverges.

The sequence converges, but the series of those terms does not.

Answer: diverges

2
Does \(\sum_{n=0}^{\infty}\dfrac{3^n}{3^n+4}\) converge or diverge?
Show Answer

Divide numerator and denominator by \(3^n\).

\(\dfrac{3^n}{3^n+4}=\dfrac{1}{1+4\cdot 3^{-n}}\)

The limit is 1, which is not 0.

The series diverges.

Answer: diverges

3
What does the nth-term test say about \(\sum_{n=1}^{\infty}\dfrac{n!}{(n+1)!}\)?
Show Answer

\(\dfrac{n!}{(n+1)!}=\dfrac{1}{n+1}\)

\(\lim_{n\to\infty}\dfrac{1}{n+1}=0\)

The test is inconclusive.

Answer: inconclusive

4
Does \(\sum_{n=1}^{\infty}\dfrac{2n+1}{5n-3}\) converge or diverge?
Show Answer

\(\lim_{n\to\infty}\dfrac{2n+1}{5n-3}=\dfrac{2}{5}\)

The limit is not 0.

The series diverges.

Answer: diverges

5
What does the nth-term test say about \(\sum_{n=2}^{\infty}\dfrac{\ln n}{n}\)?
Show Answer

\(\lim_{n\to\infty}\dfrac{\ln n}{n}=0\)

The test is inconclusive.

Answer: inconclusive

Level 3

Hard

A term that looks small, or a conclusion a student is not allowed to draw.

Example
A classmate says the terms of \(\sum \dfrac{1}{n}\) go to 0, so the series converges. What is wrong with that reason, and what is the correct conclusion?

It is true that \(1/n\to 0\).

The nth-term test does not prove convergence from that fact.

The harmonic series diverges.

The classmate’s reason is not valid, and the series diverges.

Answer: the reason is not valid, and the series diverges

1
Does \(\sum_{n=1}^{\infty}\dfrac{n^3}{n^3+n}\) converge or diverge?
Show Answer

\(\lim_{n\to\infty}\dfrac{n^3}{n^3+n}=1\)

The limit is not 0.

The series diverges.

Answer: diverges

2
What does the nth-term test say about \(\sum_{n=1}^{\infty}\left(\sqrt{n+1}-\sqrt{n}\right)\)?
Show Answer

\(\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\)

The limit is 0.

The nth-term test is inconclusive.

Answer: inconclusive

3
Does \(\sum_{n=1}^{\infty}\dfrac{5^n}{5^{n+1}}\) converge or diverge?
Show Answer

\(\dfrac{5^n}{5^{n+1}}=\dfrac{1}{5}\)

Every term equals \(\dfrac{1}{5}\).

The limit is not 0.

The series diverges.

Answer: diverges

4
Does \(\sum_{n=0}^{\infty}\cos(n\pi)\) converge or diverge?
Show Answer

\(\cos(n\pi)=(-1)^n\)

The terms alternate between 1 and \(-1\).

The limit is not 0.

The series diverges.

Answer: diverges

5
The general term of a series is \(a_n=\dfrac{4n^2-n}{n^2+6}\). A student simplifies the fraction to 4 and stops. Finish the argument.
Show Answer

The limit is 4, not the value of every term.

\(\lim a_n=4\neq 0\)

The series diverges by the nth-term test.

Answer: diverges, because the limit is 4

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