Relation for Hurwitz zeta function

Question

I would like to find a proof of the following relation for the Hurwitz zeta function $\zeta(s,a)=\sum_{n=0}^\infty \frac{1}{(n+a)^s}$: $$5\zeta(2,1/3)-\zeta(2,1/6)=\frac{4\pi^2}{3}$$

Answer 1

Let $\chi_3(3n)=0,\chi_3(3n+1)=1,\chi_3(3n+2)=-1$ and its Dirichlet L-function $L(s,\chi_3)=\sum_{n\ge 1} \chi_3(n)n^{-s}$.

$$\sum_{n=1\bmod 3} n^{-2}= \frac12(L(2,\chi_3)+(1-3^{-2})\zeta(2))$$ $$\sum_{n=1\bmod 6} n^{-2}=\frac12((1-\chi_3(2) 2^{-2}) L(2,\chi_3)+(1-2^{-2})(1-3^{-2})\zeta(2))$$ Note that $L(2,\chi_3)$ doesn't have a closed-form (similarly to $\zeta(3)$), the main point is that the terms cancel, obtaining $$5\zeta(2,1/3)-\zeta(2,1/6) = 5\cdot 3^2\sum_{n=1\bmod 3} n^{-2}-6^2 \sum_{n=1\bmod 6} n^{-2}$$ $$ = \frac{5 \cdot 3^2(1-3^{-2})-6^2(1-2^{-2})(1-3^{-2})}2 \zeta(2)=\frac43 \pi^2$$