Finding $\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right)$ Recently, I stumbled upon a summation
$$S=\sum_{n=0}^{\infty} (-1)^n\left(\frac{1}{(3n+2)^2}-\frac{1}{(3n+1)^2}\right)$$
which can luckily be summed to a good number.
Use $\psi^1(z)=\sum_{n=0}^{\infty}\frac{1}{(n+z)^2}$ to write
$$ S=\frac{1}{36}\left(\left[\psi^1\left(\frac13\right)-\psi^{1}\left(\frac56\right)\right]-\left[\psi^1\left(\frac16\right)-\psi^1\left(\frac23\right)\right]\right)$$
Next use $\psi^1(z)+\psi^1(1-z)=\pi^2\csc^2(\pi z)$ to write https://en.wikipedia.org/wiki/Polygamma_function, we get
$$I=\frac{1}{36}\left[\frac{4\pi^2}{3}-4\pi^2\right]=-\frac{2\pi^2}{27}.$$
The question is what could be other ways of getting this result?
 A: Combining consecutive terms, we have $$S = \sum_{m=0}^\infty \left(-\frac{1}{(6m+1)^2} + \frac{1}{(6m+2)^2} + \frac{1}{(6m+4)^2} - \frac{1}{(6m+5)^2}\right). \tag{1}$$
Now recall
$$\zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}, \tag{2}$$ hence
$$\sum_{m=1}^\infty \frac{1}{(3m)^2} = \frac{\zeta(2)}{9} = \frac{\pi^2}{54}. \tag{3}$$  We also have
$$\sum_{m=0}^\infty \frac{1}{(2m+1)^2} = \sum_{m=0}^\infty \frac{1}{(2m+1)^2} + \frac{1}{(2m+2)^2} - \sum_{k=1}^\infty \frac{1}{4k^2} = \frac{3}{4} \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{8}. \tag{4}$$
It follows that
$$\begin{align}
\sum_{m=0}^\infty \frac{1}{(6m+2)^2} + \frac{1}{(6m+4)^2} 
&= \frac{1}{4} \left(\!-\frac{\pi^2}{54} + \sum_{m=0}^\infty \frac{1}{(3m+1)^2} + \frac{1}{(3m+2)^2} + \frac{1}{(3m+3)^2}\! \right) \\
&= \frac{1}{4} \left( \sum_{k=1}^\infty \frac{1}{k^2} - \frac{\pi^2}{54} \right) \\
&= \frac{1}{4} \left(\frac{\pi^2}{6} - \frac{\pi^2}{54} \right) \\
&= \frac{\pi^2}{27}. \tag{5}
\end{align}$$
We apply the same trick again to $S$:
$$\begin{align}
S &= \sum_{m=0}^\infty \left(-\sum_{j=1}^6 \frac{1}{(6m+j)^2} + \frac{1}{(6m+3)^2} + \frac{1}{(6m+6)^2} \right) + 2 \frac{\pi^2}{27} \\
&= -\sum_{k=1}^\infty \frac{1}{k^2} + \sum_{m=0}^\infty \left(\frac{1}{9(2m+1)^2} + \frac{1}{36(m+1)^2} \right)+ \frac{2\pi^2}{27} \\
&= -\frac{\pi^2}{6} + \frac{\pi^2}{72} + \frac{\pi^2}{216} + \frac{2\pi^2}{27} \\
&= -\frac{2\pi^2}{27}.
\end{align}$$
