Sum of Sequence of Reciprocals of 3 n + 1 Alternating in Sign

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


 * $\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 1} = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{3 n} \rd x$

We can rewrite:

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

Specifically, we will prove that:


 * $\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x$

From Alternating Series Test: Lemma, we have:


 * $\ds \size {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \le 1$

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:


 * $\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x = \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x$

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


 * $\ds \int_0^1 \paren {\lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x = \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^3}^n} \rd x$

Then we have: