Value of $\displaystyle\sum_{k=1}^{+\infty} \log(1+\dfrac{z}{k})-\dfrac{z}{k}$

Question

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 !

Answer 1

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 $-7.6735698$

Answer 2

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?