I've been stuck with calculating the close form of series of the following problem. \begin{align*} \sum_{k=1}^{+\infty}\left(\ln(k)-\ln(x+k)+\frac{x}{k}\right) \end{align*}
for real constant $x\geq1$.
My solution: We have
\begin{align*} S_n &=\sum_{k=1}^n \Big(\ln k - \ln (k+x)+\frac{x}{n}\Big) =xH_n+\sum_{k=1}^n \ln\frac{k}{k+x}\\ &=x(\gamma+\psi_0(n+1))+\ln\prod_{k=1}^n\frac{k}{k+x}=x\gamma+x\psi_0(n+1)+\ln\frac{n!}{(x+1)(x+2)\ldots(x+n)} \end{align*} my question:
$\displaystyle{\lim_{n\to\infty}}S_n=?$
Also, I know that $\Gamma(x)=\displaystyle{\lim_{n\to\infty}}\frac{n!n^x}{x(x+1)(x+2)\ldots(x+n)}$ and so $\Gamma(x+1)=x\Gamma(x)=\displaystyle{\lim_{n\to\infty}}\frac{n!n^x}{(x+1)(x+2)\ldots(x+n)}$, but I can not $\displaystyle{\lim_{n\to\infty}}S_n$ solve. Can you help me?
We can use another approach to get the answer.
Using the Frullani integral $\,\displaystyle \ln x=\int_0^\infty\frac{e^{-x}-e^{-xt}}tdt\,$ and the fact that $\,\displaystyle\frac1k=\int_0^\infty e^{-kt}dt$ $$S(x)=\sum_{k=1}^\infty\int_0^\infty\left(\frac{e^{-t}-e^{-kt}}t-\frac{e^{-t}-e^{-(k+x)t}}t+xe^{-kt}\right)dt=\sum_{k=1}^\infty\int_0^\infty\frac{e^{-kt}}t\left(e^{-tx}+xt-1\right)dt$$ Changing the order of summation and integration and performing summation, $$S(x)=\int_0^\infty\frac{e^{-tx}+xt-1}t\frac{dt}{e^t-1}$$ $$\frac{d}{dx}S(x)=\int_0^\infty\frac{1-e^{-tx}}{e^t-1}dt\overset{s=e^{-t}}{=}\int_0^1\frac{1-s^x}{1-s}ds=\psi_0(1+x)+\gamma$$ $$\,\,\Rightarrow\,\,S(x)=\gamma x+\ln\Gamma(1+x)+C$$ Taking $x=0\,\,f(0)=0\,\,\Rightarrow\,\,C=0$ $$S(x)=\gamma x+\ln\Gamma(1+x)$$