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

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Theorem

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


Proof


Sources