Value of $\displaystyle\sum_{k=1}^{+\infty} \log(1+\dfrac{z}{k})-\dfrac{z}{k}$ I'm trying to get an explicit value of this series but I'm only able to show that it converges and that we can for $|z|\leq 1$ after expanding the log into its power series and applying Fubini's theorem we get:
\begin{align*}
\sum_{k=1}^{+\infty} \log(1+\dfrac{z}{k})-\dfrac{z}{k}&=\sum_{k=1}^{+\infty} \sum_{i=2}^{+\infty} \dfrac{(-1)^{i+1}z^i}{ik^i}
\\&=\sum_{i=2}^{+\infty}\sum_{k=1}^{+\infty}  \dfrac{(-1)^{i+1}z^i}{ik^i}
\\&=\sum_{i=2}^{+\infty}\dfrac{(-1)^{i+1}\zeta(i)}{i}z^i
\end{align*}
Does there exists an explicit form for this function ? Do we know anything about $\sum_{i=2}^{+\infty}\zeta(i)z^i$ ?
Thanks in advance !
 A: We can even copute the partial sums
$$S_p=\sum_{k=1}^p \Big[\log \left(1+\frac{z}{k}\right)-\frac{z}{k} \Big]$$ $$S_p=-z H_p-\log (\Gamma (p+1))+\zeta ^{(1,0)}(0,p+z+1)-\zeta ^{(1,0)}(0,z+1)$$ and if $p\to \infty$, then
$$\sum_{k=1}^\infty \Big[\log \left(1+\frac{z}{k}\right)-\frac{z}{k} \Big]=-\log (\Gamma (z))-\gamma  z-\log (z)$$ Using Stirling approximation, the asymptotics is
$$-z (\log (z)+\gamma -1)-\frac{1}{2} \log (2 \pi  z)-\frac{1}{12 z}+\frac{1}{360
   z^3}+O\left(\frac{1}{z^5}\right)$$
For $z=5$, the exact value is $-7.6735701$ while the asymptotics gives
$-7.6735698$
A: Hints
Some examples
\begin{align}
\displaystyle\sum_{k=1}^{+\infty} \log(1+\dfrac{1}{k})-\dfrac{1}{k}
&=-\gamma\quad\text{($\gamma$ is Euler-Mascheroni constant)}
\\
\displaystyle\sum_{k=1}^{+\infty} \log(1+\dfrac{2}{k})-\dfrac{2}{k}
&=-\ln 2 -2 \gamma
\\
\displaystyle\sum_{k=1}^{+\infty} \log(1+\dfrac{3}{k})-\dfrac{3}{k}
&=-\ln 6 -3 \gamma
\\
\displaystyle\sum_{k=1}^{+\infty} \log(1+\dfrac{4}{k})-\dfrac{4}{k}
&=-\ln 24 -4 \gamma
\\
\displaystyle\sum_{k=1}^{+\infty} \log(1+\dfrac{5}{k})-\dfrac{5}{k}
&=-\ln 120 -5 \gamma
\end{align}
Can you generalize?
