Nuestro objetivo es evaluar la suma $$\sum_{k=0}^{\infty}\left(\frac{2^{4k+1}+1}{\left(8k+1\right)}+\frac{2^{4k+2}-1}{2^{2}\left(8k+3\right)}-\frac{2^{4k+3}-1}{2^{4}\left(8k+5\right)}-\frac{2^{4k+4}+1}{2^{6}\left(8k+7\right)}\right)2^{-8k}.$$ 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 $u=\frac{x}{\sqrt{2}},$ nuestro objetivo es evaluar la 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 $x^{8}-2^{4}=\left(x^{4}-2^{2}\right)\left(x^{4}+2^{2}\right)$podemos cancelar los factores comunes y llegar a la integral
$$4\int_{0}^{1}\frac{x^{2}+2}{x^{4}+4}dx.$$ Since $x^{4}+4=\left(x^{2}-2x+2\right)\left(x^{2}+2x+2\right),$ podemos utilizar la fracción parcial de la descomposición de volver a escribir la anterior como
$$2\int_{0}^{1}\frac{1}{x^{2}-2x+2}+\frac{1}{x^{2}+2x+2}dx.$$ 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 $$2\left(\arctan\left(\frac{1}{2}\right)+\arctan\left(2\right)\right)=\pi,$$ como se desee.