Prove $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$ Just for fun:
How can we prove (calculate) that $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$ ? Can we use (9) from:
http://mathworld.wolfram.com/DigammaFunction.html
(9): $\sum_{k=0}^\infty \frac{(-1)^k}{3k+1} = \frac{1}{9}\left(\sqrt{3}\pi+3\ln\,2\right)$
?
Thx!


We can evaluate the series of interest without appealing to the Digamma Function.

Note that we can simply write
$$\begin{align}
\sum_{n=0}^{2N}\frac{(-1)^{n}}{3n+2}&=\sum_{n=0}^N\left(\frac{1}{6n+2}-\frac{1}{6n+5}\right)\\\\
&=\sum_{n=0}^N\int_0^1\left( x^{6n+1}-x^{6n+6}\right)\,dx\\\\
&=\int_0^1 x\left(\frac{1-x^{6N+6}}{1+x^3}\right)\,dx\\\\
&=\int_0^1 \left(\frac{x}{1+x^3}\right)\,dx -\int_0^1 x^{6N+7} \left(\frac{1}{1+x^3}\right)\,dx
\end{align}$$
Applying the Dominated Convergence Theorem (or alternatively, integrate by parts and observe that the second integral is $ O(N^{-1})$), we find that
$$\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+2}=\int_0^1 \frac{x}{1+x^3}\,dx$$
Can you finish now using partial fraction expansion for example?
