Is there an easy way to show that $F(z)=\sum_{j=1}^{z}\zeta(j+1)-\zeta(z+5)-z$?

Question

$$\sum_{n=1}^{\infty}\left[\zeta(2n+z)-2\zeta(n+z+1)+\zeta(n+z+2)-\zeta(n+z+3)+\zeta(2n+z+5)\right]=F(z)$$

Is there is a simple way to show that ($z\ge0$) $$F(z)=\sum_{j=1}^{z}\zeta(j+1)-\zeta(z+5)-z$$

Answer 1

Let $f_n$ be a sequence such that $\sum_{0<n}f_n$ absolutely converge, then \begin{align} &\sum_{0<n}(f_{2n}-2f_{n+1}+f_{n+2}-f_{n+3}+f_{2n+5})\\ &=\sum_{0<n}(f_{2n}+f_{2n+5})-2\sum_{0<n}f_{n+1}+\sum_{0<n}(f_{n+2}-f_{n+3})\\ &=\sum_{0<n}(f_{2n}+f_{2n+1})-\sum_{0<n}(f_{2n+1}-f_{2n+5})-2\sum_{0<n}f_{n+1}-f_3\\ &=\sum_{0<n}f_{n+1}-(f_3+f_5)-2\sum_{0<n}f_{n+1}-f_3\\ &=-\sum_{0<n}f_{n+1}-f_5 \end{align} Put $f_n=\zeta(n+z)-1$, we have \begin{align} F(z)&=\sum_{0<n}(\zeta(2n+z)-2\zeta(n+z+1)+\zeta(n+z+2)-\zeta(n+z+3)+\zeta(2n+z+5))\\ &=-\sum_{0<n}(\zeta(n+z+1)-1)-(\zeta(z+5)-1)\\ &=-\sum_{0<n}(\zeta(n+1)-1)+\sum_{j=1}^z(\zeta(j+1)-1)-\zeta(z+5)+1\\ &=1-\sum_{0<n}(\zeta(n+1)-1)+\sum_{j=1}^z(\zeta(j+1)-1)-\zeta(z+5)-z \end{align} And, $$ \sum_{0<n}(\zeta(n+1)-1)=1 $$ is easy to show.