Finding sum of Maclaurin series Find sum of the series $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+3)(2n+1)}.$$
I think I have to find known taylor series and modify it to look like the above but I can't see which one to pick.
The solution is $\frac{\pi -2}{4}$ so I'm guessing it has to be one of the trig functions. 
I picked $\sin x$
$$\sin(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$
integrate
$$-\cos(x)=\sum_{n=0}^\infty \frac{(-1)^{n} x^{2n+2}}{(2n+2)(2n+1)!}.$$
The index is not correct plus I don't think I can get it to look like the problem.
Can someone explain which function to pick and how to solve?
 A: $$\begin{array}{rcl}
\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1}
&=& \displaystyle \sum_{n=0}^\infty \int_0^1 (-1)^n x^{2n} \ \mathrm dx \\
&=& \displaystyle \sum_{n=0}^\infty \int_0^1 (-x^2)^n \ \mathrm dx \\
&=& \displaystyle \int_0^1 \sum_{n=0}^\infty (-x^2)^n \ \mathrm dx \\
&=& \displaystyle \int_0^1 \dfrac1{1+x^2} \ \mathrm dx \\
&=& \displaystyle \left(\arctan x\right)_0^1 \\
&=& \displaystyle \dfrac\pi4 \\
\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)(2n+3)}
&=& \dfrac12 \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} - \dfrac12\displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+3} \\
&=& \dfrac12 \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} + \dfrac12\displaystyle \sum_{n=1}^\infty \dfrac{(-1)^n}{2n+1} \\
&=& \displaystyle \sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1} - \dfrac12 \\
&=& \dfrac\pi4 - \dfrac12 \\
\end{array}$$
A: Maybe you will like this method. Let
$$ f(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+3)(2n+1)}x^{2n+3}.$$
Then
$$ f'(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+2}, (\frac{f(x)}{x})'=\sum_{n=0}^\infty(-1)^nx^{2n}=\frac{1}{1+x^2}.$$
So
$$ f(1)=\int_0^1x\int_0^x\frac{1}{1+t^2}dtdx=\int_0^1x\arctan xdx=\frac{\pi}{4}-\frac12. $$
