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

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.
