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

Proof
From Primitive of Power and the Fundamental Theorem of Calculus, we have:


 * $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {4 n + 1} = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{4 n} \rd x$

We can rewrite:

We now use Lebesgue's Dominated Convergence Theorem to swap integral and summation.

Specifically, we will prove that:


 * $\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x$

From Alternating Series Test: Lemma, we have:


 * $\ds \size {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \le 1$

for all $x \in \closedint 0 1$ and $N \in \N$.

All functions involved are continuous, so the conditions for Lebesgue's Dominated Convergence Theorem are satisfied, and we have:


 * $\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x$

Then, by the definition of infinite series, we have:


 * $\ds \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{4 n} } \rd x = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^4}^n} \rd x$

Then we have:

We have:

and:

Putting these together gives:


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