Finding the Sum of an Infinite Series with a Given Radius |x|<1

[stefaneli]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

I've tried to do the following:

S&#039;(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n}}{2n-1} \\<br /> S&#039;&#039;(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{2nx^{2n-1}}{2n-1}\\<br /> S&#039;&#039;&#039;(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.

 

[tiny-tim]

hi stefaneli!

S&#039;(x) = \sum_{n=1}^{∞}(-1)^{n+1}\frac{x^{2n}}{2n-1}

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

 

[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.

 

[tiny-tim]

… can it be done the way I started?

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

 

[tt2348]

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

 

[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