Evaluating $\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$ Evaluate : $$\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$$
I tried using logarithmic functions, but cannot evaluate this please help me
 A: To apply the DFT (discrete Fourier transform) to the Taylor series of $\log(1+x)$ is a perfectly viable way, but the simplest approach is probably just to write $\frac{1}{3n+1}$ as $\int_{0}^{1}x^{3n}\,dx$, then noticing that
$$ \sum_{n\geq 1}\frac{(-1)^n}{3n+1} = \int_{0}^{1}\sum_{n\geq 0}(-x)^{3n}\,dx =\int_{0}^{1}\frac{dx}{1+x^3}\stackrel{\text{PFD}}{=}\color{red}{\frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}}$$
where $\text{PFD}$ stands for partial fraction decomposition.
A: HINT:
Consider $\frac{1}{1+x^3} = \sum(-1)^nx^{3n}$
A: Since you know that
$\ln(1+x)
=\sum_{n=1}^{\infty} x^n/n
$ for $-1\le x < 1$,
look up multisection of series
to evaluate your sum.
A: One could make use of the digamma function and obtain:
\begin{align}
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{x \, k + 1} &= \frac{1}{2 x} \, \left[ \psi\left(\frac{x+1}{2 x}\right) - \psi\left(\frac{1}{2 x}\right) \right] \\
\psi\left(\frac{1}{6}\right) &= - \gamma - \frac{\sqrt{3} \, \pi}{2} - \frac{3}{2} \, \ln(3) - 2 \, \ln(2) \\
\psi\left(\frac{2}{3}\right) &= - \gamma - \frac{\pi}{2 \sqrt{3}} - \frac{3}{2} \, \ln(2).
\end{align}
Setting $x = 3$ yields the desired result.
