1. ## Series

1.) find the series' radius and interval of convergence. for what values of x does the series converge absolutely and conditionally?

Σ [(3^n)(x^n)] / (n!)
n=0

2.) find the sum of the series by expressing 1/ (1-x) as a geometric series

Σ (n^2)/(2^n)
n=0

2. 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.