1. ## Series equation

Could anyone help me out with this equation please?

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

How to prove it? Any ideas?

2. ## Re: Series equation

Yes consider the partial fraction expansion of $\displaystyle \frac{1}{n^3-n}$.

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

P.S. join mathhelpboards.com - this one is done for.

3. ## Re: Series equation

Could you write it out please? You mean $\frac{1}{n(n-1)(n+1)}$?

4. ## Re: Series equation

Yes, try to write it as

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

Deduce what A, B, and C are.
Do you know how to?
Then split the summation into 3 summations.

5. ## Re: Series equation

OK, I did it. $A = -1, B = \frac {1}{2}, C = \frac{1}{2}$. Is it ok?

6. ## Re: Series equation

Originally Posted by Lisa91
OK, I did it. $A = -1, B = \frac {1}{2}, C = \frac{1}{2}$. Is it ok?
Yes.

7. ## Re: Series equation

So, I got something like this: $-\frac{2}{n3^{n}} + \frac{1}{(n-1)3^{n}} + \frac{1}{(n+1)3^{n}}$ I still don't see what I can take out of this...

8. ## Re: 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.

9. ## Re: Series equation

Oh sorry stupid question.

10. ## Re: Series equation

Originally Posted by Lisa91
Oh sorry stupid question.
If you didn't understand that then write out the terms of the series and cancel what you can cancel.

11. ## Re: Series equation

You can split the summation into 3 summations.

The second summation is:

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

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

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

12. ## Re: Series equation

The third term is $\sum_{m=3}^{\infty} \frac{3}{m3^{m}}$. How can I 'go back' to 1?

13. ## Re: Series equation

By subtracting the terms at m = 2, and m = 1.

$\sum_{m=1}^{\infty} \frac{3}{m3^{m}} -\frac{3}{m3^{m}}\bigg|_{m=1}-\frac{3}{m3^{m}}\bigg|_{m=2}$

14. ## Re: Series equation

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

15. ## Re: Series equation

3S+S/3-2S Surely you can combine that.

Make all indexes k if that's confusing you.

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