If the series $\displaystyle\sum_{n=1}^\infty a_n$ converges, then $\displaystyle\lim_{n\to\infty}a_n = 0$. This is the same as saying that if $\displaystyle\lim_{n\to\infty}a_n \ne 0$ (or the limit doesn’t exist), then the series $\displaystyle\sum_{n=1}^\infty a_n$ diverges.

• Example: $\displaystyle\sum_{n=1}^\infty(-2)^n$ diverges because $\displaystyle\lim_{n\to\infty}(-2)^n \ne 0$.

Note: having $\displaystyle\lim_{n\to\infty}a_n = 0$ is not enough to know that $\displaystyle\sum_{n=1}^\infty a_n$ converges!

• Example: $\displaystyle\sum_{n=1}^\infty\frac{1}{n} = \infty$, but $\displaystyle\lim_{n\to\infty}\frac{1}{n} = 0$.

an example involving factorials

$$ \sum_{n=1}^\infty\left[\frac{n^2(2n-1)!}{(2n+1)!}\right] $$

Let’s focus on the $\frac{(2n-1)!}{(2n+1)!}$ term. We can write this as:

$$ \frac{(2n-1)!}{(2n+1)!} = \frac{(2n-1)(2n)\cdots(2)(1)}{(2n+1)(2n)(2n-1)(2n)\cdots(2)(1)} = \frac{1}{(2n+1)(2n)} $$

Therefore, the sum can be written as:

$$ \sum_{n=1}^\infty\left[\frac{n^2(2n-1)!}{(2n+1)!}\right] = \sum_{n=1}^\infty\left[\frac{n^2}{(2n+1)(2n)}\right] =\sum_{n=1}^\infty\left[\frac{n^2}{4n^2+2n}\right] $$

We can use the $n$th-term test here to conclude that this series diverges:

$$ \lim_{n\to\infty}\frac{n^2}{4n^2+2n} = \frac{1}{4}\ne0 $$