Proving that $\pi=\sum\limits_{k=0}^{\infty}(-1)^{k}\left(\frac{2^{2k+1}+(-1)^{k}}{(4k+1)2^{4k}}+ \frac{2^{2k+2}+(-1)^{k+1}}{(4k+3)2^{4k+2}}\right)$

Our goal is to evaluate the sum $$\sum_{k=0}^{\infty}\left(\frac{2^{4k+1}+1}{\left(8k+1\right)}+\frac{2^{4k+2}-1}{2^{2}\left(8k+3\right)}-\frac{2^{4k+3}-1}{2^{4}\left(8k+5\right)}-\frac{2^{4k+4}+1}{2^{6}\left(8k+7\right)}\right)2^{-8k}.$$ We split this into two different convergent sums, $$\sum_{k=0}^{\infty}\left(\frac{1}{\left(8k+1\right)}-\frac{1}{2^{2}\left(8k+3\right)}+\frac{1}{2^{4}\left(8k+5\right)}-\frac{1}{2^{6}\left(8k+7\right)}\right)2^{-8k}\ \ \ \ \ (1)$$ and $$\sum_{k=0}^{\infty}\left(\frac{2^{4k+1}}{\left(8k+1\right)}+\frac{2^{4k+2}}{2^{2}\left(8k+3\right)}-\frac{2^{4k+3}}{2^{4}\left(8k+5\right)}-\frac{2^{4k+4}}{2^{6}\left(8k+7\right)}\right)2^{-8k}.\ \ \ \ \ (2)$$ The sum $(1)$ is easily seen to be $$2\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}\left(\frac{1}{2}\right)^{2k+1}=2\arctan\left(\frac{1}{2}\right).$$ For the second sum, we may rearrange the terms and write it as $$\left(\sqrt{2}\right)^{3}\sum_{k=0}^{\infty}\left(\frac{1}{\left(8k+1\right)(\sqrt{2})^{8k+1}}+\frac{1}{\left(8k+3\right)(\sqrt{2})^{8k+3}}-\frac{1}{\left(8k+5\right)\left(\sqrt{2}\right)^{8k+5}}-\frac{1}{\left(8k+7\right)\left(\sqrt{2}\right)^{8k+7}}\right).$$ By considering generating functions of the form $\sum_{k=0}^{\infty}\frac{1}{8k+1}\left(x\right)^{8k+1}$, we see that the above is equal to $$\left(\sqrt{2}\right)^{3}\int_{0}^{1/\sqrt{2}}\frac{1+u^{2}-u^{4}-u^{6}}{1-u^{8}}du.$$ Making the substitution $u=\frac{x}{\sqrt{2}},$ our goal is to evaluate the integral

$$4\int_{0}^{1}\frac{x^{6}+2x^{4}-4x^{2}-8}{x^{8}-2^{4}}du.$$ Factoring the numerator as $x^{6}+2x^{4}-4x^{2}-8=(x^{2}-2)(x^{2}+2)^{2}$, and the denominator as $x^{8}-2^{4}=\left(x^{4}-2^{2}\right)\left(x^{4}+2^{2}\right)$, we may cancel the common factors and arrive at the integral

$$4\int_{0}^{1}\frac{x^{2}+2}{x^{4}+4}dx.$$ Since $x^{4}+4=\left(x^{2}-2x+2\right)\left(x^{2}+2x+2\right),$ we may use the partial fraction decomposition to rewrite the above as

$$2\int_{0}^{1}\frac{1}{x^{2}-2x+2}+\frac{1}{x^{2}+2x+2}dx.$$ Factoring the denominators as $(x-1)^2+1$ and $(x+1)^2+1$, we see that this is in fact twice the integral of $\frac{1}{x^{2}+1}$ from $0$ to $2$, which is $2\arctan(2)$. Thus, adding the results for series $(1)$ and series $(2)$ together, we find that the original sum equals $$2\left(\arctan\left(\frac{1}{2}\right)+\arctan\left(2\right)\right)=\pi,$$ as desired.

See the update for the answer to question. Below follows a proof that the series are the same.

Here is how I did it: Note that every positive integer can be written either as an even or odd number (of the form $2k$ for $k \in \mathbb{N}$ and $2k+1$ respectively). Then, since your sum goes through all the integers, it is the same as: $$ \begin{align} &\sum_{k=0}^{\infty} \left(\frac{2^{4k+1}+1}{(8k+1)}+\frac{2^{4k+2}-1}{(8k+3)2^{2}}-\frac{2^{4k+3}-1}{(8k+5)2^{4}}-\frac{2^{4k+4}+1}{(8k+7)2^{6}} \right)\frac{1}{2^{8k}}\\ &=\sum\limits_{k=0}^{\infty}(-1)^{2k}\ \left(\frac{2^{4k+1}+(-1)^{2k}}{(8k+1)2^{8k}}+ \frac{2^{4k+2}+(-1)^{2k+1}}{(8k+3)2^{8k+2}} \right)\\ &+\sum\limits_{k=0}^{\infty}(-1)^{2k+1} \left(\frac{2^{4k+3}+(-1)^{2k+1}}{(8k+5)2^{8k+2}}+ \frac{2^{4k+4}+(-1)^{2k+2}}{(8k+7)2^{8k+3}}\right) \end{align} $$

The two sum expressions are simply your original sum, but with $2k$ respectively $2k+1$ inserted instead of k.

Update

I misunderstood the question. How to derive this series should be easy to find here: http://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula