# Riemann Zeta function and Catalan's constant

I'm struggling at figuring out this identity: $$G:=\beta(2)=\frac{1}{16}\sum_{n=1}^{\infty}(n+1)\frac{3^n-1}{4^n}\zeta(n+2).$$ with $$G$$ the Catalan's constant, $$\zeta$$ the Riemann Zeta function and $$\beta$$ the Dirichlet beta function.

The identity I understand is $$\beta(s)=4^{-s}\left(\zeta(s,\frac{1}{4})-\zeta(s,\frac{3}{4})\right).$$ with $$\zeta(s,q)$$ being the Hurwitz Zeta function. This result is quite easy to prove, acknowledging that \begin{align} \zeta(s,\frac{1}{4})-\zeta(s,\frac{3}{4})&= \sum_{n=0}^{\infty} \left(\frac{1}{(n+\frac{1}{4})^s}-\frac{1}{(n+\frac{3}{4})^s}\right)=\sum_{n=0}^{\infty} 4^s\left(\frac{1}{(4n+1)^s}-\frac{1}{(4n+3)^s}\right)\\ &=4^s \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^s}. \end{align} However, I don't really see how plugging in $$s=2$$ yields the result from above, namely how $$\zeta(2,\frac{1}{4})-\zeta(2,\frac{3}{4})=\sum_{n=1}^{\infty}(n+1)\frac{3^n-1}{4^n}\zeta(n+2).$$ Thanks a lot in advance!

• Commented May 7, 2020 at 22:13

$$\begin{eqnarray} \zeta(s,q) &=& \sum_{k=0}^\infty\frac1{(k+q)^s} \\ &=& \sum_{k=1}^\infty\frac1{(k+q-1)^s} \\ &=& \sum_{k=1}^\infty\frac1{k^s}\cdot\frac1{\left(1+\frac{q-1}k\right)^s} \\ &=& \sum_{k=1}^\infty\frac1{k^s}\sum_{n=0}^\infty\binom{-s}n\left(\frac{q-1}k\right)^n \\ &=& \sum_{n=0}^\infty\binom{-s}n(q-1)^n\sum_{k=1}^\infty\frac1{k^{n+s}} \\ &=& \sum_{n=0}^\infty\binom{-s}n(q-1)^n\zeta(n+s)\;. \end{eqnarray}$$
In your case, $$\binom{-2}n=(-1)^n(n+1)$$, so
$$\begin{eqnarray} \zeta\left(2,\frac14\right)-\zeta\left(2,\frac34\right) &=& \sum_{n=0}^\infty\binom{-2}n\left(\left(\frac14-1\right)^n-\left(\frac34-1\right)^n\right)\zeta(n+2) \\ &=& \sum_{n=0}^\infty(n+1)\left(\left(\frac34\right)^n-\left(\frac14\right)^n\right)\zeta(n+2) \;. \end{eqnarray}$$
• I think it has something to do with the fact that $\zeta(s,q)$ starts at $k=0$ per definition. But then the first equality is problematic because one would divide through $0$. Commented May 7, 2020 at 21:55