Sum of Sequence of Reciprocals of 4 n + 1 Alternating in Sign

Proof
We have:

and:

Putting these together gives:


 * $\displaystyle \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} = \int_0^1 \frac 1 {1 + x^4} \rd x = \frac {\pi \sqrt 2} 8 + \frac {\sqrt 2 \map \ln {1 + \sqrt 2} } 4$