Find the sum of the series: $\frac{1}{1*2} - \frac{1}{3*2^3} + \frac{1}{5*2^5} - \frac{1}{7*2^7}+\dots$? $$\frac{1}{1*2} - \frac{1}{3*2^3} + \frac{1}{5*2^5} - \frac{1}{7*2^7}+\dots$$ 
I made a series to get $$\sum_{n=0}^\inf \frac{(-1)^n}{(1+2n)*2^{1+2n}}$$ but what series can it manipulate and simplify to?
 A: The geometric series gives $$\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \,\,\,\,\,\,\,\, \text{when } \lvert x \rvert < 1.$$ Then $$\frac{1}{1+x^2} - 1 = \sum_{n=1}^{\infty} (-1)^n x^{2n},$$ for such $x$. Integrating from $x=0$ to $x=1/2$ (and integrating term-by-term in the sum, which is justifies by uniform convergence of the sum) gives $$\arctan\left( \tfrac 1 2\right) - \tfrac{1}{2} = \sum^\infty_{n=1} \frac{(-1)^n \left( \tfrac{1}{2}\right)^{2n+1}}{2n+1} = \sum^\infty_{n=1} \frac{(-1)^n}{(2n+1)2^{2n+1}}.$$
A: Let $f(x)$ denote the power series for $|x|<1$ given by 
$$\begin{align}
f(x)&=\sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{2n+1}\\\\
&=\sum_{n=0}^\infty (-1)^n\int_0^x t^{2n}\,dt\\\\
&=\lim_{N\to \infty}\int_0^x \sum_{n=0}^N (-t^2)^n \,dt\\\\
&=\lim_{N\to \infty}\int_0^x \frac{1-(-t^2)^{N+1}}{1+t^2}\,dt \tag 1 \\\\
&=\int_0^x\frac{1}{1+t^2}\,dt \tag 2\\\\
&=\arctan(x)
\end{align}$$
where the Dominated Convergence Theorem justifies our passing the limit in $(1)$ under the integral and arriving at $(2)$.  Now, setting $x=1/2$, we obtain the coveted result
$$f(1/2)=\arctan(1/2)$$
