How to evaluate $\sum_{n=2}^{\infty}\frac{\zeta (n)-1}{n-1}$ and $\sum_{n=2}^{\infty}\frac{(-1)^{n}\left ( \zeta(n)-1 \right )}{n-1}$? I am interested in the rational zeta series as title. WolframAlpha gives the result:
$$
\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1}=0.78853056591150896106... 
$$
This is the Lüroth analog of Khintchine’s constant, which is defined as:
$$
\sum_{n=1}^{\infty}\frac{\ln (n)}{n(n+1)}=-\sum_{n=2}^{\infty}(-1)^{n}\zeta^{'}(n)
$$
Why this rational zeta series is not in terms of $\gamma$, $\ln2$, $\ln \pi$ or normal constant that usually seen, but comes out an unusual constant? How to evaluate it to a closed form?
Also how to evaluate its companion zeta series to a closed form?
$$
\sum_{n=2}^{\infty}\frac{(-1)^{n}\left ( \zeta(n)-1 \right )}{n-1}=0.56459970638442432059... 
$$
Or there are no closed forms for both zeta series?
 A: Although there are no closed forms for both rational zeta series, we can evaluate them with definite integrals.
Set $f(x)$ as
\begin{align*}
f(x)&=\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1}x^{n-1}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{k^n}\int_{0}^{x}t^{n-2}\,dt=\int_{0}^{x}\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{t^{n-2}}{k^n}\,dt \\
&=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\sum_{n=2}^{\infty}\left( \frac{t}{k} \right)^{n-2}dt=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\frac{1}{\left(1-\frac{t}{k}\right)}\,dt \\
&=\int_{0}^{x}\left(\frac{1}{t-1}+\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{(k-t)} \right)dt=\int_{0}^{x}\frac{1}{t-1}\,dt-\int_{0}^{x}\frac{\gamma+\psi(1-t)}{t}\,dt \\
&=\ln(1-x)+\left [ \frac{\ln\Gamma(1-t))}{t} \right ]_{t=0}^{t=x}+\int_{0}^{x}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\
&=\ln(1-x)+\frac{\ln\Gamma(1-x)}{x}-\gamma+\int_{0}^{x}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\ \\
f(1)&=\sum_{n=2}^{\infty}\frac{\zeta(n)-1}{n-1} \\
&=\lim_{x\rightarrow 1^{-}}\left ( \ln(1-x))+\frac{\ln\Gamma(1-x))}{x} \right )-\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \\
&=-\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1-t)}{t^2}-\frac{\gamma}t{} \right )dt \approx0.7885306 \\
\end{align*}
In the same way, set $f(x)$ as
\begin{align*}
f(x)&=\sum_{n=2}^{\infty}\frac{(-1)^{n}(\zeta(n)-1)}{n-1}x^{n-1}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{(-1)^{n}}{k^n}\int_{0}^{x}t^{n-2}\,dt=\int_{0}^{x}\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{(-t)^{n-2}}{k^n}\,dt \\
&=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\sum_{n=2}^{\infty}\left( -\frac{t}{k} \right)^{n-2}dt=\int_{0}^{x}\sum_{k=2}^{\infty}\frac{1}{k^2}\frac{1}{\left(1+\frac{t}{k}\right)}\,dt \\
&=\int_{0}^{x}\left(-\frac{1}{t+1}+\sum_{k=1}^{\infty}\frac{1}{k}\frac{1}{(k+t)} \right)dt=-\int_{0}^{x}\frac{1}{t+1}\,dt+\int_{0}^{x}\frac{\gamma+\psi(1+t)}{t}\,dt \\
&=-\ln(1+x)+\left [ \frac{\ln\Gamma(1+t))}{t} \right ]_{t=0}^{t=x}+\int_{0}^{x}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \\
&=-\ln(1+x)+\frac{\ln\Gamma(1+x)}{x}+\gamma+\int_{0}^{x}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \\ \\
f(1)&=\sum_{n=2}^{\infty}\frac{(-1)^{n}(\zeta(n)-1)}{n-1} \\
&= -\ln(2)+\gamma+\int_{0}^{1}\left ( \frac{\ln\Gamma(1+t)}{t^2}+\frac{\gamma}t{} \right )dt \approx0.5645997 \\
\end{align*}
