<s>[INST] Prove that $\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$ 
Prove that $$\sum_{n=0}^\infty \frac{(-1)^n}{3n+1} = \frac{\pi}{3\sqrt{3}}+\frac{\log 2}{3}$$

I tried to look at $$ f_n(x) = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^n $$
And maybe taking it's derivative but it didn't work out well.
Any ideas?
 [/INST] For a geometric series.
$$\sum_{n=0}^{\infty} (-1)^n(x^n) = \frac{1}{1+x}$$
Substitute $x \rightarrow x^3$
$$\sum_{n=0}^{\infty} (-1)^n (x^{3n}) = \frac{1}{1+x^3}$$
Integrate the sides.
$$\sum_{n=0}^{\infty} (-1)^n\frac{x^{3n + 1}}{3n + 1} = \int \frac{dx}{1+x^3}$$
The hard part is the integration. Then let $x=1$
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