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

1. Jul 3, 2012

### Hernaner28

1. The problem statement, all variables and given/known data

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

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

2. Relevant equations

3. 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

2. Jul 3, 2012

### 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?

3. Jul 3, 2012

### 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?

4. Jul 3, 2012

### 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?

5. Jul 3, 2012

### 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

6. Jul 3, 2012

### Hernaner28

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