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

[Hernaner28]

Homework Statement

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

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

Homework Equations

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

 

[tt2348]

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?

 

[Hernaner28]

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?

 

[tt2348]

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?

 

[tt2348]

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

 

[Hernaner28]

Wow, that's pretty tough, THANK YOUU!