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

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Theorem

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 1}\) \(=\) \(\ds 1 - \frac 1 4 + \frac 1 7 - \frac 1 {10} + \frac 1 {13} - \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\pi \sqrt 3} 9 + \dfrac {\ln 2} 3\)


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:

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{3 n} \rd x\) \(=\) \(\ds \lim_{N \mathop \to \infty} \sum_{n \mathop = 0}^N \paren {-1}^n \int_0^1 x^{3 n} \rd x\) Definition of Infinite Series
\(\ds \) \(=\) \(\ds \lim_{N \mathop \to \infty} \int_0^1 \paren {\sum_{n \mathop = 0}^N \paren {-1}^n x^{3 n} } \rd x\) Linear Combination of Definite Integrals

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:

\(\ds \int_0^1 \paren {\sum_{n \mathop = 0}^\infty \paren {-x^3}^n} \rd x\) \(=\) \(\ds \int_0^1 \frac 1 {1 + x^3} \rd x\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \intlimits {\frac 1 6 \ln \size {\frac {\paren {x + 1}^2} {x^2 - x + 1} } + \frac 1 {\sqrt 3} \map \arctan {\frac {2 x - 1} {\sqrt 3} } } 0 1\) Primitive of $\dfrac 1 {x^3 + a^3}$
\(\ds \) \(=\) \(\ds \frac 1 6 \ln 4 + \frac 1 {\sqrt 3} \map \arctan {\frac 1 {\sqrt 3} } - \frac 1 6 \ln 1 - \frac 1 {\sqrt 3} \map \arctan {-\frac 1 {\sqrt 3} }\)
\(\ds \) \(=\) \(\ds \frac 1 3 \ln 2 + \frac 2 {\sqrt 3} \map \arctan {\frac 1 {\sqrt 3} }\) Logarithm of Power, Arctangent Function is Odd, Natural Logarithm of 1 is 0
\(\ds \) \(=\) \(\ds \frac 1 3 \ln 2 + \frac 2 {\sqrt 3} \times \frac \pi 6\) Arctangent of $\dfrac {\sqrt 3} 3$
\(\ds \) \(=\) \(\ds \frac {\pi \sqrt 3} 9 + \frac {\ln 2} 3\)

$\blacksquare$


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