Series equation

**Lisa91** · January 15th 2013, 08:36 AM

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?

**TheSaviour** · January 15th 2013, 08:47 AM

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.

**Lisa91** · January 15th 2013, 09:31 AM

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

**ILikeSerena** · January 15th 2013, 09:47 AM

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.

**Lisa91** · January 15th 2013, 09:58 AM

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

**TheSaviour** · January 15th 2013, 10:00 AM

Originally Posted by Lisa91

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

Yes.

**Lisa91** · January 15th 2013, 10:03 AM

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...

**ILikeSerena** · January 15th 2013, 10:06 AM

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.

**Lisa91** · January 15th 2013, 10:09 AM

Oh sorry stupid question.

**TheSaviour** · January 15th 2013, 10:12 AM

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.

**ILikeSerena** · January 15th 2013, 10:13 AM

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:

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

**Lisa91** · January 15th 2013, 10:23 AM

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

**TheSaviour** · January 15th 2013, 10:33 AM

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

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

**Lisa91** · January 15th 2013, 10:41 AM

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...

**TheSaviour** · January 15th 2013, 10:51 AM

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

Make all indexes k if that's confusing you.

Thread: Series equation

LinkBack

Thread Tools

Display

Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Re: Series equation

Similar Math Help Forum Discussions

Series ( tricky equation )

A Product Series Equation

Regression of series equation

solve differential equation by series

Differential equation and power series

Search Tags