Too many things to handle!

The sum is $\begin{array}{rcl} S & = & \displaystyle 2\sum_{k=1}^\infty \frac{\zeta(k+1)-1}{(k+1)(k+2)} \; = \; 2\sum_{k=2}^\infty \frac{\zeta(k)-1}{k(k+1)} \; = \; 2\sum_{k=2}^\infty \frac{\zeta(k)-1}{k} - 2\sum_{k=2}^\infty \frac{\zeta(k)-1}{k+1} \\ & = & \displaystyle 2 - 2\gamma - 2\sum_{k=2}^\infty \frac{\zeta(k)-1}{k+1} \end{array}$ and $\begin{array}{rcl} \displaystyle \sum_{k=2}^\infty \frac{\zeta(k)-1}{k+1} & = & \displaystyle \sum_{k=2}^\infty \sum_{n=2}^\infty \frac{1}{(k+1)n^k} \; = \; \sum_{n=2}^\infty \sum_{k=2}^\infty \frac{1}{(k+1)n^k} \\ & = & -\displaystyle \sum_{n=2}^\infty \left[ n \ln\big(1- n^{-1}\big) + 1 + \tfrac12n\right] \; = \; -\lim_{N \to \infty} X_N \;, \end{array}$ where $\begin{array}{rcl} X_N & = & \displaystyle \sum_{n=2}^N \left[ n \ln(n-1) - n \ln n + 1 + \tfrac12n\right] \\ & = & \displaystyle \sum_{n=1}^{N-1} (n+1) \ln n - \sum_{n=2}^N n \ln n + (N-1) + \tfrac12(H_N - 1) \\ & = & \displaystyle \sum_{N=1}^{N-1} \ln n - N \ln N + N + \tfrac12H_N - \tfrac32 \\ & = & \displaystyle \ln N! - (N+\tfrac12)\ln N + N + \tfrac12(H_N - \ln N) - \tfrac32 \\ & = & \displaystyle \ln\left(\frac{N! e^N}{N^{N+\frac12}}\right) + \tfrac12(H_N - \ln N) - \tfrac32 \\ & \to & \ln \sqrt{2\pi} + \tfrac12\gamma - \tfrac32 \;, \end{array}$ as $N \to \infty$ , using Stirling's approximation. Thus $S \; = \; 2 - 2\gamma + 2\ln\sqrt{2\pi} + \gamma - 3 \; = \; \ln(2\pi) - \gamma - 1$ making the answer $1 + 2 + 1 + 1 = \boxed{5}$ .