Evaluating $\sum_{n=0}^\infty\frac{(-1)^n}{3n+1}$ [closed]

Solution 1:

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.

Solution 2:

HINT:

Consider $\frac{1}{1+x^3} = \sum(-1)^nx^{3n}$

Solution 3:

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.