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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?

Kyan Cheung
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    Could be expressed in terms of the di-gamma function, as I see it. – Mourad Sep 11 '20 at 04:29
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    Note that $H_x$ is only defined for positive integer values. But the di-gamma function can be treated as an analytic continuation of this. and we have $$ψ\left(x+1\right)=H_{x}-γ$$ Check out the wiki page: https://en.wikipedia.org/wiki/Digamma_function – Mourad Sep 11 '20 at 04:32
  • @Mourad I could replace $H_x$ with $\psi(x+1)+\gamma$, but how would I prove that this is the correct answer for non-integer $x$? – Kyan Cheung Sep 11 '20 at 04:38
  • According to Wikipedia $\psi(x+1)+\gamma=\int^1_0\frac{1-t^x}{1-t}dt$, maybe that could be useful. – Kyan Cheung Sep 11 '20 at 04:40
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    It involves taking this limit-definition of the gamma function $$Γ\left(x+1\right)=\lim_{n \rightarrow \infty}\frac{n!n^{x}}{\left(x+1\right)\left(x+2\right)..\left(x+n\right)}$$ And differentiating it logarithmic ally. From there, you'll reach your desired sum. – Mourad Sep 11 '20 at 04:48
  • Not sure if that helps, but using Euler's partial fraction series for the cotangent function, you can get something like $S(x)+S(-x)=\frac 1 {x^2}-\frac {\pi}{x}\cot(\pi x)$ for $x$ not an integer. – Stefan Lafon Sep 11 '20 at 04:57
  • @Mourad Thanks for the hint, I was able to figure out how to show $\sum^{\infty}_{n=1}\frac1{n(n+x)}=\frac{\psi(x+1)+\gamma}{x}$. – Kyan Cheung Sep 11 '20 at 05:10

3 Answers3

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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)

Kyan Cheung
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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 $x\in\Bbb R^+\setminus\Bbb N$.

J.G.
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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 $$S_p=\frac{H_x}{x}-\frac{1}{p}+\frac{x+1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$

Claude Leibovici
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  • I wasn't looking for $\sum^p_{n=1}\frac1{n(n+x)}$ for finite $p$. I was looking at the infinite sum for non-integer $x$, and I've already figured out what it equals to (see my answer). – Kyan Cheung Sep 11 '20 at 06:54
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    @Kyky. I know that. I made the sum up to $p$, used asymptotics and made $p\to \infty$. This is a classic procedure. Moreover, my limit is shorter than your even if identical. Cheers :-) – Claude Leibovici Sep 11 '20 at 07:15
  • I see, thanks for the clarification. – Kyan Cheung Sep 11 '20 at 07:16