Series

A handy test in this area is the n-th root test: given a series $\sum_n a_n$ , for which the sequence $a_n^{1/n}$ tends to a limit $\ell$ , the series is convergent if $\ell < 1$ and divergent if $\ell > 1$ . (In the case $\ell=1$ , the test gives no information.)

Applied to your series (1), we see that $(3^n x^n / n!)^{1/n} = 3x/(n!)^{1/n}$ and it isn't too hard to show that $(n!)^{1/n} \rightarrow \infty$ . Hence the limit $\ell = 0$ and the series is convergent for any value of x.

For your series (2) a useful trick is to regard it as a special value of a function $f(x) = \sum_n n^2 x^n$ , evaluated at $x=1/2$ . We start with $1/(1-x) = \sum_n x^n$ and differentiate to get $1/(1-x)^2 = \sum_n n x^{n-1}$ . Multiply by x and differentiate again to find f.