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.
 A: In the language of Dirichlet series and the Riemann zeta function I believe this could be counted as an elementary proof:
Add the variable $s$ as an exponent to your series so that it becomes:
$$\sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s}+\frac{1}{(6n+5)^s}+\frac{2}{(6n+6)^s}\right)$$
$$=\zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)
$$
In the case of $s=1$ we have exactly your series.
Therefore we investigate the limit:
$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)$$
taking only parts of the limit we have:
$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)=\log(2)$$
and:
$$\lim_{s\to 1} \, \left(1-\frac{1}{3^{s-1}}\right)=0$$
therefore we have:
$$\lim_{s\to 1} \, \zeta(s)\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)=\log(2) \cdot 0 = 0$$
hence:
$$\lim_{s\to 1} \, \sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s}+\frac{1}{(6n+5)^s}+\frac{2}{(6n+6)^s}\right)=0$$
which is equivalent to:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
A: 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.
A: I think we can "squeeze" something out of this:
$$0=\sum_{n=0}^\infty\left(\frac{1}{6n+6}+\frac{-1}{6n+6}+\frac{-2}{6n+6}+\frac{-1}{6n+6}+\frac{1}{6n+6}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)\le\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+1}+\frac{-2}{6n+1}+\frac{-1}{6n+1}+\frac{1}{6n+1}+\frac{2}{6n+1}\right)=0$$
A: This is a comment in response to Mats Granvik's answer. 
I bet Mats Granvik's Trick generalizes to all Weinberger series!
Let $$f(s,\vec{a})=\sum_{n=1}^\infty {\frac{a_n}{n^s}} $$ 
be called a Weinberger series when $f(1,\vec{a})=0$
where $\vec{a}=(a_1, a_2,\dots a_p)$ and $\forall k\in \mathbb{N}, \text{ }a_{p+k}=a_{k}$. 
Let $p=q_1^{e_1} \dots q_l^{e_l}$ be the prime factorization of $p$. 
Conjecture $$f(1,\vec{a})=0\implies f(s,\vec{a})=\lim_{x\to s}\zeta(x)\prod_{i=1}^l(1-\frac{1}{q_i^{x-1}})^{e_i}$$
Let's consider another case. How about  $\vec{a}=[1,-3,1,1]$?
$f(\vec{a},s)=\zeta(s)(1-\frac{1}{2^{s-1}})^2$ 
I think generally speaking the factor-ability of $f$ may rely on the series being a Weinberger series. 
By the way this would mean that $p$ prime $\implies f(1, \vec{a}) \neq 0$  and I think this is the case based on proposition 13 of this paper. 
A: Write term $n$ in the form
$$
\left(\frac1{6n+1}+\frac1{6n+2}+\cdots+\frac1{6n+6}\right)
-\left(\frac1{3n+1}+\frac1{3n+2}+\frac1{3n+3}\right)
-\left(\frac1{2n+1}+\frac1{2n+2}\right)+\frac1{n+1}.
$$ Sum from $n=0$ to $n=N-1$. This gives the $N$th partial sum for the series:
$$\left(\sum_1^{6N}\frac1k-\sum_1^{3N}\frac1k\right)
-\left(\sum_1^{2N}\frac1k-\sum_1^N\frac1k\right)=\sum_{3N+1}^{6N}\frac1k-\sum_{N+1}^{2N}\frac1k.
$$
Let $N\to\infty$. Each sum on the RHS converges to $\log 2$, by this result.
