# How can I calculate the sum of the following infinite series?

## Homework Statement

1) $$\displaystyle \sum\limits_{n=1}^{\infty }{\frac{{{n}^{3}}}{{{e}^{n}}}}$$

2) $$\displaystyle \sum\limits_{n=1}^{\infty }{\frac{-1}{2\left( n+1 \right)}+\frac{2}{\left( n+2 \right)}-\frac{3}{n\left( n+3 \right)}}$$

## The Attempt at a Solution

I've proved that both series converge. In fact the second series was given like this:
$$\displaystyle \sum\limits_{n=1}^{\infty }{\frac{n}{\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}}$$
I just simplified to calculate the sum. Anyway, I don't know how to do that. I don't know when I am able to split the series into sum of series because I know that it cannot be done sometimes but sometimes it can.

Thanks

for
$\sum\limits_{n=1}^{\infty}\frac{n}{(n+1)(n+2)(n+3)}$
I get when expanded
$\sum\limits_{n=1}^{\infty}\frac{n}{(n+1)(n+2)(n+3)}=\sum\limits_{n=1}^{\infty}-\frac{1}{2}(\frac{1}{n+1})+2(\frac{1}{n+2})-\frac{3}{2}(\frac{1}{n+3})=\sum\limits_{n=1}^{ \infty} \frac{1}{2}(\frac{1}{n+2}-\frac{1}{n+1})+\frac{3}{2}(\frac{1}{n+2}-\frac{1}{n+3})$
any ideas what to do with that?

Yeah, it's a telescopic. It was really simple. Didn't realize.

And for the first, any idea?

Thanks!

BTW Do you know in which cases it's possible to separate the series into sum of series?

for telescoping... when you see things like n/(n+k)!, or 1/(n+1)(n+2).. style
it's just an intuitive feeling you'll learn.

for the first.. I have a method, but it's not pretty,
are you familiar with power series?

for x<1
$f(x)=\sum\limits_{n=1}^{\infty}x^{n}=\frac{1}{1-x}$
then $f'(x)=\sum\limits_{n=1}^{\infty}nx^{n-1}$
implies $xf'(x)=\sum\limits_{n=1}^{\infty}nx^{n}$
so $x\frac{d}{dx}(xf'(x))=\sum\limits_{n=1}^{\infty}n^2x^{n}$
and finally $g(x)=x\frac{d}{dx}(x\frac{d}{dx}(xf'(x)))=\sum \limits_{n=1}^{ \infty}n^3x^{n}$

using f(x)=1/(1-x)... , find g(x), and plug in 1/e

Wow, that's pretty tough, THANK YOUU!!!