Calculus/Limit Test for Convergence

Limit Test

The first test for divergence is the limit test. The limit test essentially tells us whether or not the series is a candidate for being convergent. It is as follows:

Limit Test for Convergence

If a series $S=\sum _{n}^{\infty }{s_{n}}$ and if $\lim _{n\rightarrow \infty }{s_{n}}\neq 0$ the series must be divergent. If the limit is zero, the test is inconclusive and further analysis is needed.

This follows because if the summand does not approach zero, when $n$ becomes very large, $s_{n}$ will be close to the non-zero $L$ and the series will start behaving like an arithmetic series; remember that arithmetic series never converge.

However, one should not misuse this test. This is a test for divergence and not convergence. A series fails this test if the limit of the summand is zero, not if it is some non-zero $L$. If the limit is zero, you will need to do other tests to conclude that the series is divergent or convergent.

Example 1

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n+1}{n}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n+1}{n}}=1$

the limit is not zero and so the series is divergent by the limit test.

Example 2

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{3n^{2}+1}}={\frac {1}{3}}$

the limit is not zero and so the series is divergent by the limit test.

Example 3

Determine whether or not the series

$\sum _{n=0}^{\infty }{\frac {n^{2}+5n+6}{n^{3}}}$

is divergent or if the limit test fails.

Solution

Because

$\lim _{n\rightarrow \infty }{\frac {n^{2}+5n+6}{n^{3}}}=0$

the limit is zero and so the test is inconclusive. Further analysis is needed to determine whether or not the series converges or diverges.