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.

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I believe all of those "well known" harmonic-like sums have been proven using calculus. If so, then why do you expect there to be a non-calculus solution to evaluate this sum? –  JimmyK4542 Jul 31 at 2:14
Because the answer is 0. I feel that such a simple answer deserves a simple proof. –  columbus8myhw Jul 31 at 2:14
There is, I"m typing it up right now. –  Semiclassical Jul 31 at 2:16
^Ahh, very well done. –  JimmyK4542 Jul 31 at 2:23

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.

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Ah. I wonder why I didn't notice that before! Good job. –  columbus8myhw Jul 31 at 2:24
Good answer. $\log(2)-\log(2)=0$ –  robjohn Jul 31 at 2:26

I think we can "squeeze" something out of this: $$0=\sum_{n=0}^\infty\left(\frac{1}{6n+6}+\frac{-1}{6n+6}+\frac{-2}{6n+6}+\frac{-1}{6n+6}+\frac{1}{6n+6}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+1}+\frac{-2}{6n+1}+\frac{-1}{6n+1}+\frac{1}{6n+1}+\frac{2}{6n+1}\right)=0$$

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Ah. Never mind. The negative terms in the sum would reverse the inequalities for those terms. This will not work. –  Laars Helenius Jul 31 at 7:43
The previous answer gets my up vote then. –  Laars Helenius Jul 31 at 7:45