Considering the Odd & Even Sequences of Gamma & Zeta Limits: $$ \begin{align} & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)(2n+2)}}\space\quad=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-1)}{2}-\frac{\Gamma(x+1)\zeta(x+1)}{2!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+3)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-2)}{2}-\frac{\Gamma(x-1)\zeta(x-1)}{1!}-\frac{\Gamma(x+1)\zeta(x+1)}{3!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+4)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-3)}{2}-\frac{\Gamma(x-1)\zeta(x-1)}{2!}-\frac{\Gamma(x+1)\zeta(x+1)}{4!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+5)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-4)}{2}-\frac{\Gamma(x-3)\zeta(x-3)}{1!}-\cdots-\frac{\Gamma(x+1)\zeta(x+1)}{5!}\right]\,, \\[2mm] & \small\color{blue}{\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(2n+1)\cdots(2n+6)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-5)}{2}-\frac{\Gamma(x-3)\zeta(x-3)}{2!}-\cdots-\frac{\Gamma(x+1)\zeta(x+1)}{6!}\right]\,, \quad\cdots \\[2mm] & {\bf=} \space\color{blue}{\frac{1}{2}}-\frac{\gamma}{2}{\bf\,,} \quad\color{blue}{\frac{7}{24}}-\frac{\gamma}{6}+\frac{\zeta'(-1)}{1}{\bf\,,} \quad\color{blue}{\frac{1}{9}}-\frac{\gamma}{24}+\frac{\zeta'(-1)}{2}{\bf\,,} \\[2mm] & \quad\color{blue}{\frac{277}{8640}}-\frac{\gamma}{120}+\frac{\zeta'(-1)}{6}+\frac{\zeta'(-3)}{6}{\bf\,,} \quad\color{blue}{\frac{79}{10800}}-\frac{\gamma}{720}+\frac{\zeta'(-1)}{24}+\frac{\zeta'(-3)}{12}{\bf\,,} \quad\cdots \\[2mm] & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align} $$
$$ \begin{align} & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)(2n+1)}}\space\quad=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-1)}{2}+\frac{\Gamma(x+0)\zeta(x+0)}{1!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+2)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-2)}{2}+\frac{\Gamma(x+0)\zeta(x+0)}{2!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+3)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-3)}{2}+\frac{\Gamma(x-2)\zeta(x-2)}{1!}+\frac{\Gamma(x+0)\zeta(x+0)}{3!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+4)}}=\lim_{x\rightarrow0}\left[+\frac{\Gamma(x-4)}{2}+\frac{\Gamma(x-2)\zeta(x-2)}{2!}+\frac{\Gamma(x+0)\zeta(x+0)}{4!}\right]\,, \\[2mm] & \small\color{red}{\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)\cdots(2n+5)}}=\lim_{x\rightarrow0}\left[-\frac{\Gamma(x-5)}{2}+\frac{\Gamma(x-4)\zeta(x-4)}{1!}+\cdots+\frac{\Gamma(x+0)\zeta(x+0)}{5!}\right]\,, \quad\cdots \\[2mm] & {\bf=} \space\color{red}{\frac{1}{2}}+\frac{\zeta'(0)}{1}{\bf\,,} \quad\color{red}{\frac{3}{8}}+\frac{\zeta'(0)}{2}{\bf\,,} \quad\color{red}{\frac{11}{72}}+\frac{\zeta'(0)}{6}+\frac{\zeta'(-2)}{2}{\bf\,,} \\[2mm] & \quad\color{red}{\frac{25}{576}}+\frac{\zeta'(0)}{24}+\frac{\zeta'(-2)}{4}{\bf\,,} \quad\color{red}{\frac{137}{14400}}+\frac{\zeta'(0)}{120}+\frac{\zeta'(-2)}{12}+\frac{\zeta'(-4)}{24}{\bf\,,} \quad\cdots \\[2mm] & \quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\color{white}{\text{.}} \end{align} $$
Find a formula to calculate either (the first term) or (the limit itself) in each case?
$ SO_{n}=\left\{\frac{1}{2}\,,\,\frac{7}{24}\,,\,\frac{1}{9}\,,\,\frac{277}{8640}\,,\,\frac{79}{10800}\,,\,\cdots\right\} \quad\&\quad SE_{n}=\left\{\frac{1}{2}\,,\,\frac{3}{8}\,,\,\frac{11}{72}\,,\,\frac{25}{576}\,,\,\frac{137}{14400}\,,\,\cdots\right\} $
Give $N\ge1\rightarrow$ Find $\small\{\color{blue}{SO_{N}}\,\&\,\color{red}{SE_{N}}\}$ OR $\small\{\color{blue}{\sum\frac{\zeta(2n+1)}{(2n+1)(2n+2)\cdots(2n+N+1)}}\,\&\,\color{red}{\sum\frac{\zeta(2n)}{(2n)(2n+1)\cdots(2n+N)}}\}$