Sum of Sequence of Reciprocals of 3 n + 2 Alternating in Sign

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Theorem

\(\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 2}\) \(=\) \(\ds \frac 1 2 - \frac 1 5 + \frac 1 8 - \frac 1 {11} + \frac 1 {14} - \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\pi \sqrt 3} 9 - \dfrac {\ln 2} 3\)


Proof

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

$\ds \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac 1 {3 n + 2} = \sum_{n \mathop = 0}^\infty \paren {-1}^n \int_0^1 x^{3 n + 1} \rd x$

We can rewrite:

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

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

Specifically, we will prove that:

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

From Alternating Series Test: Lemma, we have:

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

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

So:

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

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

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

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

Then we have:

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

$\blacksquare$


Sources

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