Originally Posted by

**parallel** hello

I need to find the sum:

sigma 0-->infinity (-1/3)^n * 1/(2n+1)

i looked at the power series : x^2n (1)

the sum of this series is : 1/(1-x^2) (2)

then by integration of (1) I got : x^2n+1 * 1/(2n+1) (3)

by integration of (2) i got the sum of (3)

I dont know what to do now

help would be appreciated.

Consider the series expansion:

$\displaystyle

\arctan(x)=\sum_0^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1},\ \ |x|<1

$,

so:

$\displaystyle

\arctan(\sqrt{x})=\sum_0^{\infty}\frac{(-1)^n x^{(2n+1)/2}}{2n+1}$

$\displaystyle

=x^{1/2}\sum_0^{\infty}\frac{(-x)^{n}}{2n+1},\ \ |x|<1

$

and the rest should follow easily.

RonL