The Easy Peasy Sum of Series

Define the function $\displaystyle f(x)= \sum_{n=0}^{\infty} \frac{(-1)^nx^{4n+1}}{4^{4n+1}(4n+1)}$ , where we seek $f(1)$ and $f(0)=0$ .

$\displaystyle f'(x)= \sum_{n=0}^{\infty} \frac{(-1)^nx^{4n}}{4^{4n+1}}= \frac{1}{4} \sum_{n=0}^{\infty} (-\frac{x^4}{256})^n = \frac{1}{4} \frac{1}{1+\frac{x^4}{256}}$ (Geometric Series)

$\displaystyle => f(1)=f(1)-f(0)=\int_0^1 f'(x)dx= \int_0^1 \frac{1}{4} \frac{1}{1+\frac{x^4}{256}} dx$

$\displaystyle = \int_0^{\frac{1}{4}} \frac{du}{u^4+1} (u=\frac{x}{4})$ .

Now this integral can be solved using partial fractions

( $x^4+1=(x^2+1)^2-2x^2=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$ )

To get at last the value $\displaystyle \frac{\ln \frac{17+4\sqrt2}{17-4\sqrt2} + 2\arctan \frac{4+\sqrt2}{4} -2\arctan \frac{4-\sqrt2}{4}}{4\sqrt2} \approx \boxed{0.25}$

integrate summation(x^4n*(-1)^n , from 0 to inf)