# Sum of infinite series

1. Jun 4, 2012

### stefaneli

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

I need to find the sum of a given infinite series when $|x|<1$ (which is the radius of this series)

2. Relevant equations

$\sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n+1}}{4n^2-1}$

3. The attempt at a solution
I've tried to do the following:

$S'(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n}}{2n-1} \\ S''(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{2nx^{2n-1}}{2n-1}\\ S'''(x) = 2\sum_{n=1}^{∞}(-1)^{n+1}nx^{2(n-1)}\\$

And I was thinking about substitution $t = x^2$, but I had no success.

2. Jun 4, 2012

### tiny-tim

hi stefaneli!
$= x\sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n-1}}{2n-1}$

3. Jun 4, 2012

### stefaneli

Thanks tiny-tim. I don't know how I haven't noticed. Stupid.:) But can it be done the way I started? Just curious.

4. Jun 4, 2012

### tiny-tim

i suppose so, but you'd need to be very careful when you change the variable

5. Jun 4, 2012

### tt2348

Try doing partial fractions. IE 1/(4n^2-1)= 1/2*(1/(2n-1)-1/(2n+1))

6. Jun 4, 2012

### tt2348

Your answer will be something like 1/2*((x^2-1)arctan(x)-x), just use partial fractions, and get it into a more manageable form