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

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  • #1
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Homework Statement



1) [tex] \displaystyle \sum\limits_{n=1}^{\infty }{\frac{{{n}^{3}}}{{{e}^{n}}}}[/tex]

2) [tex] \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)}}[/tex]


Homework Equations





The Attempt at a Solution



I've proved that both series converge. In fact the second series was given like this:
[tex] \displaystyle \sum\limits_{n=1}^{\infty }{\frac{n}{\left( n+1 \right)\left( n+2 \right)\left( n+3 \right)}}[/tex]
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
 

Answers and Replies

  • #2
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for
[itex]\sum\limits_{n=1}^{\infty}\frac{n}{(n+1)(n+2)(n+3)}[/itex]
I get when expanded
[itex]\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})[/itex]
any ideas what to do with that?
 
  • #3
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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
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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
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for x<1
[itex]f(x)=\sum\limits_{n=1}^{\infty}x^{n}=\frac{1}{1-x} [/itex]
then [itex]f'(x)=\sum\limits_{n=1}^{\infty}nx^{n-1} [/itex]
implies [itex]xf'(x)=\sum\limits_{n=1}^{\infty}nx^{n} [/itex]
so [itex]x\frac{d}{dx}(xf'(x))=\sum\limits_{n=1}^{\infty}n^2x^{n}[/itex]
and finally [itex] g(x)=x\frac{d}{dx}(x\frac{d}{dx}(xf'(x)))=\sum \limits_{n=1}^{ \infty}n^3x^{n}[/itex]

using f(x)=1/(1-x)... , find g(x), and plug in 1/e
 
  • #6
263
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Wow, that's pretty tough, THANK YOUU!!!
 

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