What is $\sum^{\infty}_{n=1}\frac1{n(n+x)}$ equal to? I was wondering how we could calculate the sum $S(x)=\sum^{\infty}_{n=1}\frac1{n(n+x)}$ for any real $x$.
I've noted the following properties regarding the sum (which may or may not be useful to actually finding $S(x)$):

*

*We have the identity $\sum^{\infty}_{n=1}\frac1{n(n+x)}=\frac1x\sum^{\infty}_{n=1}\frac1{n}-\frac1{n+x}$ for $x\neq0$. If we rearrange the terms in the sum, we find that for $x\in\mathbb{Z}_+$, $S(x)=\frac{H_x}{x}$, where $H_x$ is the xth harmonic number. This means $S(x)\sim\frac{\ln x}x$ for positive $x$.

*The case $x=0$ is  the Basel Problem, so we know $S(0)=\frac{\pi^2}6$.

*For $x\in\mathbb{Z}_-$, there is an $n$ such that $n+x=0$ and $\frac1{n(n+x)}=\pm\infty$. So $S(x)=\pm\infty$ for $x\in\mathbb{Z}_-$.

*$S'(x)=-\sum^{\infty}_{n=1}\frac1{n(n+x)^2}$. As $\frac1{n(n+x)^2}$ is negative, this means that (ignoring discontinuities) $S(x)$ is strictly decreasing.

However, I have no idea how to actually get a closed form of $\sum^{\infty}_{n=1}\frac1{n(n+x)}$ for non-integer $x$. How could I calculate this sum?
 A: One of the definitions of the gamma function is that
$$\Gamma(x+1)=\lim_{n\to\infty}\frac{n!x^n}{(x+1)(x+2)\cdots(x+n)}$$
Working from here,
$$\ln(\Gamma(x+1))=\lim_{n\to\infty}\ln(n!)+x\ln n-\ln(x+1)-\ln(x+2)-\cdots-\ln(x+n)$$
$$\frac d{dx}\ln(\Gamma(x+1))=\lim_{n\to\infty}\ln n-\frac1{x+1}-\frac1{x+2}-\cdots-\frac1{x+n}$$
$$\psi(x+1)=\lim_{n\to\infty}\ln n-\sum^n_{i=1}\frac1{x+i}$$
$$\psi(x+1)=\lim_{n\to\infty}H_n+\left(\ln n-H_n\right)-\sum^n_{i=1}\frac1{x+i}$$
$$\psi(x+1)=\lim_{n\to\infty}\sum^n_{i=1}\frac1{i}+\left(\ln n-\ln n-\gamma-O\left(\frac1n\right)\right)-\sum^n_{i=1}\frac1{x+i}$$
$$\psi(x+1)=\lim_{n\to\infty}\sum^n_{i=1}\left[\frac1{i}-\frac1{x+i}\right]-\gamma-O\left(\frac1n\right)$$
$$\psi(x+1)=\sum^\infty_{i=1}\left[\frac1{i}-\frac1{x+i}\right]-\gamma$$
$$\sum^\infty_{i=1}\left[\frac1{i}-\frac1{x+i}\right]=\psi(x+1)+\gamma$$
$$\frac1x\sum^\infty_{i=1}\left[\frac1{i}-\frac1{x+i}\right]=\frac{\psi(x+1)+\gamma}x$$
$$\sum^\infty_{i=1}\frac1{i(i+x)}=\frac{\psi(x+1)+\gamma}x$$
Thus $$S(x)=\frac{\psi(x+1)+\gamma}x$$
(Credit to @Mourad for providing hints for me)
A: Your sum is$$\frac1x\sum_{n\ge1}\int_0^1t^{n-1}(1-t^x)dt=\frac1x\int_0^1\frac{1-t^x}{1-t}dt=\frac{H_x}{x}.$$With the integral-based definition of continuous Harmonic numbers, this is valid even for $x\in\Bbb R^+\setminus\Bbb N$.
A: As you wrote it
$$S_p=\frac 1x \sum_{n=1}^p\left(\frac{1}{n }-\frac{1}{n+x} \right)=\frac 1x\left(H_p +H_x-H_{p+x}\right)$$
Using the asymptotics of harmonic numbers
$$S_p=\frac{H_x}{x}-\frac{1}{p}+\frac{x+1}{2
   p^2}+O\left(\frac{1}{p^3}\right)$$
