Determine which of the following series, , converge
i)
ii)
$\displaystyle a_n = \frac {n^2}{n!} = \frac {n}{(n-1)!}$
Using the Ratio test:
$\displaystyle \lim_{n->\infty} \left| \frac{a_{n+1}}{a_n}\right| = \lim_{n->\infty} \left| \frac{n+1}{n!}*\frac{(n-1)!}{n}\right|$
$\displaystyle =\lim_{n->\infty} \left| \frac{n+1}{n^2}\right|$
$\displaystyle =\lim_{n->\infty} \left| \frac{1+\frac 1n}{n}\right|$
$\displaystyle =0$
0 < 1 therefore the series is absolutely convergent and therefore convergent.
Not sure if that was a joke, but it was rather amusing ^_^