A certain “harmonic” sum

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.

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: $$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$