<s>[INST] Weird infinite sum Evaluate
$\sum_{n=0}^{\infty}{\frac{(-1)^n}{3n+1}=1-\frac{1}{4}+\frac{1}{7}-\frac{1}{10}+...}
$
This looks a lot like the series expansion for $\ln(1+x)$ when $x=1$, but I cannot find the relationship.
 [/INST] The log series hints at finding the function
$$ f(x) = \sum_{n=0}^{\infty} \frac{(-1)^nx^{3n+1}}{3n+1} $$
when $x=1$. Taking the derivative, we find 
$$ f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{3n} = \sum_{n=0}^{\infty} (-x^3)^n = \frac{1}{1+x^3} $$
Therefore
$$ f(1) = \int_0^1 \frac{1}{1+t^3} dt $$
You can solve the integral using partial fractions. WolframAlpha gives the solution as well as the anti-derivative here
$$ f(x) = \frac13 \ln (x+1) - \frac16 \ln(x^2-x+1) + \frac{1}{\sqrt{3}}\arctan \left(\frac{2x-1}{\sqrt{3}}\right) $$
$$ f(1) = \frac13 \ln 2 + \frac{\pi}{3\sqrt{3}} $$
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