Could anyone help me out with this equation please?

$\displaystyle \sum_{n=2}^{\infty} \frac{2}{(n^{3} - n)3^{n}} = - \frac{1}{2} + \frac{4}{3} \sum_{n=1}^{\infty} \frac {1}{n3^{n}}$

How to prove it? Any ideas?

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- Jan 15th 2013, 08:36 AMLisa91Series equation
Could anyone help me out with this equation please?

$\displaystyle \sum_{n=2}^{\infty} \frac{2}{(n^{3} - n)3^{n}} = - \frac{1}{2} + \frac{4}{3} \sum_{n=1}^{\infty} \frac {1}{n3^{n}}$

How to prove it? Any ideas? - Jan 15th 2013, 08:47 AMTheSaviourRe: Series equation
Yes consider the partial fraction expansion of $\displaystyle \displaystyle \frac{1}{n^3-n}$.

Two of the fractions will telescope and you'll be done.

P.S. join mathhelpboards.com - this one is done for. ;) - Jan 15th 2013, 09:31 AMLisa91Re: Series equation
Could you write it out please? You mean $\displaystyle \frac{1}{n(n-1)(n+1)} $?

- Jan 15th 2013, 09:47 AMILikeSerenaRe: Series equation
Yes, try to write it as

$\displaystyle {A \over n} + {B \over n - 1} + {C \over n + 1}$

Deduce what A, B, and C are.

Do you know how to?

Then split the summation into 3 summations. - Jan 15th 2013, 09:58 AMLisa91Re: Series equation
OK, I did it. $\displaystyle A = -1, B = \frac {1}{2}, C = \frac{1}{2} $. Is it ok?

- Jan 15th 2013, 10:00 AMTheSaviourRe: Series equation
- Jan 15th 2013, 10:03 AMLisa91Re: Series equation
So, I got something like this: $\displaystyle -\frac{2}{n3^{n}} + \frac{1}{(n-1)3^{n}} + \frac{1}{(n+1)3^{n}} $ I still don't see what I can take out of this...

- Jan 15th 2013, 10:06 AMILikeSerenaRe: Series equation
You have a summation that starts at n=2.

Rewrite the summation of the second term by replacing n-1 by for instance k.

And rewrite the summation of the third term by replacing n+1 by m.

Afterward add or subtract whatever you need to, to make the summations start at 1.

Then you can combine them again. - Jan 15th 2013, 10:09 AMLisa91Re: Series equation
Oh sorry stupid question.

- Jan 15th 2013, 10:12 AMTheSaviourRe: Series equation
- Jan 15th 2013, 10:13 AMILikeSerenaRe: Series equation
You can split the summation into 3 summations.

The second summation is:

$\displaystyle \sum_{n=2}^\infty \frac 1 {(n-1)3^n}$

Substitute n=k+1 and you'll get:

$\displaystyle \sum_{k+1=2}^\infty \frac 1 {(k+1-1)3^{k+1}} = \sum_{k=1}^\infty \frac 1 {k 3^{k+1}} = \frac 1 3 \sum_{n=1}^\infty \frac 1 {n 3^n}$ - Jan 15th 2013, 10:23 AMLisa91Re: Series equation
The third term is $\displaystyle \sum_{m=3}^{\infty} \frac{3}{m3^{m}} $. How can I 'go back' to 1?

- Jan 15th 2013, 10:33 AMTheSaviourRe: Series equation
By subtracting the terms at m = 2, and m = 1.

$\displaystyle \sum_{m=1}^{\infty} \frac{3}{m3^{m}} -\frac{3}{m3^{m}}\bigg|_{m=1}-\frac{3}{m3^{m}}\bigg|_{m=2}$ - Jan 15th 2013, 10:41 AMLisa91Re: Series equation
So, I got $\displaystyle \sum_{m=1}^{\infty} \frac{3}{m3^{m}} - \frac{7}{6} + \frac{1}{3}\sum_{k=1}^{\infty} \frac{1}{k3^{k}} - \sum_{n=1}^{\infty} \frac{2}{n3^{n}} $ How to combine them all? I don't know how to cope with those $\displaystyle 2,3, \frac{1}{3} $. For instance, if I go back to n, I get n=0 in the first term, which is not what I want...

- Jan 15th 2013, 10:51 AMTheSaviourRe: Series equation
3S+S/3-2S Surely you can combine that. :D

Make all indexes k if that's confusing you.