A certain “harmonic” sum Is there a simple, elementary proof of the fact that:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for example:
$[\overline{1}]=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\dots=\infty\;$ is the harmonic series
$[\overline{1,-1}]=\frac{1}{1}+\frac{-1}{2}+\frac{1}{3}+\dots=\ln2\;$ is well known
$[\overline{1,1,-2}]=\frac{1}{1}+\frac{1}{2}+\frac{-2}{3}+\dots=\ln3\;$ is slightly less well known (I think)
$[\overline{1,0,-1,0}]=\frac{1}{1}+\frac{0}{2}+\frac{-1}{3}+\dots=\frac{\pi}{4}\;$ is the Gregory-Leibniz series for $\pi$
What I claim is that $[\overline{1,-1,-2,-1,1,2}]$ is equal to $0$. I wonder if there are any simple proofs of this (i.e. definitely without using calculus, preferably without appealing to complex numbers/taylor series/etc.)
P.S. I know a method that doesn't use any integrals or derivatives, but requires knowledge of the taylor series for $\ln(x)$ and the Euler formula for $e^{ix}$.
The reason I believe that there should be an elementary proof is that the sum, $0$, is a very simple number.

We may rewrite your series in the following manner:
\begin{align}
&\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\\
&=\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{1}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{-1}{6n+6}\right)\\
&\hspace{1cm}-\sum_{n=0}^\infty\left(\frac{3}{6n+3}-\frac{3}{6n+6}\right)\\
\end{align}
But these summations are both the alternating series $\sum_{n=0}^\infty \dfrac{(-1)^n}{n+1}$. Therefore they cancel and the summation is equal to zero.
