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.
 A: 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
$$ f(x) = \frac13 \ln (x+1) - \frac16 \ln(x^2-x+1) + \frac{1}{\sqrt{3}}\arctan \left(\frac{2x-1}{\sqrt{3}}\right) $$
$$ f(1) = \frac13 \ln 2 + \frac{\pi}{3\sqrt{3}} $$
A: We will use imprecise argument to guess the value of the sum. We know that
$$\sum_{n=0}^\infty x^{3n} = \frac{1}{1-x^3}.$$
By integrating it both sides we have $$\sum_{n=0}^\infty \frac{x^{3n+1}}{3n+1} = \int_0^x \frac{1}{(1-u)(1+u+u^2)}du = \frac{1}{6}\left(\log\left(\frac{x^2+x+1}{(1-x)^2}\right)+ 2\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right).$$
Hence we can guess
$$\sum_{n=0}^\infty \frac{(-1)^{3n+1}}{3n+1} = \frac{1}{6}\left(\log\left(\frac{1}{4}\right)- \frac{2\sqrt{3}\pi}{6}\right).$$
Our argument is not strict as we need to justify we can exchange the integral and the infinite sum. Moreover, we don't know our sum is coincide with the result of the integration.
(Note. The sum in left-hand-side is not equal to your sum, but it is in fact same as the negation of your sum.)
We can justisfy our result as follows: note that 
$$\frac{1}{1-x^3} = \sum_{n=0}^N x^{3n} + \frac{x^{3N+3}}{1-x^3}$$
We can integrate from 0 to -1 it both sides then we have
$$\frac{1}{6}\left(\log\left(\frac{1}{4}\right)- \frac{2\sqrt{3}\pi}{6}\right) = \sum_{n=0}^N \frac{-1}{3n+1} + \int_0^{-1}\frac{x^{3N+3}}{1-x^3}dx$$
Here the last term can be regarded as an errer term. It satisfies
$$
\begin{align}
\left|\int_0^{-1}\frac{x^{3N+3}}{1-x^3}dx\right|& 
\le \int_{-1}^0\left|\frac{x^{3N+3}}{1-x^3}\right|dx \\
&\le \int_{-1}^0 |x|^{3N+3}dx\\
&= \frac{1}{3N+4} \xrightarrow[N\to\infty]{} 0,
\end{align}$$
since $|1/(1-x^3)|\le 1$ if $-1\le x\le 0$. Now take $N\to\infty$.
A: Recall how we can obtain the Taylor series of $\ln (1+x)$ - Integrate $\frac{1}{1+x} = 1-x+x^{2}-x^{3}+\cdots$ term by term. (You should be careful for $x=1$, but it still works by Abel's limit theorem.) Now your series maybe obtained by 
$$
\int_{0}^{1}\frac{1}{1+x^{3}}dx.
$$
(Also, there's some convergence issue for $x=1$. But certainly we can deal with it.)
