A double sum involving the Riemann zeta function Evaluate the sum $S=\sum_{k=2}^{\infty} \frac{\zeta (k)-1}{k+1}$, where $\zeta (s)$ denotes the Riemann zeta function.
The sum is equal to $\sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{(k+1)n^k}$, then switching the order (since the summand converges uniformly) gives $$S=\sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{(k+1)n^k}$$.
For the inner sum, I tried to evaluate it by considering $\sum_{k=0}^{\infty} x^k = \frac{1}{1-x}$ for $|x|<1$. Integrating w.r.t $x$ and then divided both sides by $x$ gives
$$\sum_{k=2}^{\infty}  \frac{x^k}{k+1}= \frac{-1}{x}\ln |1-x|-1-\frac{x}{2}$$. Then put $x=1/n$ for $n=2,3,\cdots$. But then I dont how to evaluate the outer sum, please helps.

Remember Stirling's formula
$$\sum_{m=1}^n \log m = \log (n!) = \left(n+\tfrac{1}{2}\right)\log n - n + \tfrac{1}{2}\log (2\pi) + O\left(\tfrac{1}{n}\right).$$
Then we have
$$\begin{align}
\sum_{n=2}^K \left(-n\log\left(1 - \tfrac{1}{n}\right) - 1 - \tfrac{1}{2n}\right)
&= 1-K +\tfrac{1}{2} - \tfrac{1}{2}H_K - \sum_{n=2}^K n\log(n-1) + \sum_{n=2}^K n\log n\\
&= \tfrac{3}{2} - K - \tfrac{1}{2}H_K - \sum_{n=2}^{K-1}\log n + K\log K\\
&= \left(\left(K+\tfrac{1}{2}\right)\log K - K + \tfrac{1}{2}\log (2\pi) - \sum_{n=1}^K\log n\right)\\
&\qquad + \tfrac{3}{2}+ \tfrac{1}{2}\underbrace{\left(\log K - H_K\right)}_{-\gamma + O(K^{-1})} - \tfrac{1}{2}\log (2\pi)\\
&= \tfrac{3}{2} - \tfrac{1}{2}\gamma - \tfrac{1}{2}\log (2\pi) +  O\left(\tfrac{1}{K}\right),
\end{align}$$
where $H_K$ is the $K$-th harmonic number.
