# Asymptotic behavior of Harmonic-like series $\sum_{n=1}^{k} \psi(n) \psi'(n) - (\ln k)^2/2$ as $k \to \infty$

I would like to obtain a closed form for the following limit: $$I_2=\lim_{k\to \infty} \left ( - (\ln k)^2/2 +\sum_{n=1}^{k} \psi(n) \, \psi'(n) \right)$$

Here $\psi(n)$ is digamma function.

Using the method detailed in this answer, I was able to compute simpler, related series: $$\lim_{k\to \infty} \sum_{n=1}^{k} \left (\psi'(n) -1/n \right) =1$$ $$\sum_{n=1}^{\infty} \psi''(n) =-\frac{\pi^2}{3}$$ But $I_2$ seems to be tougher because of the product of two digamma's. The divergence of $(\ln k)^2$ is matched by the first terms of asymptotic series for $\psi(n) \psi'(n)$, via the definition of the Stieltjes number, $$\lim_{k\to \infty} \left ( \sum_{n=1}^{k} \frac{\ln n}{n} - (\ln k)^2/2 \right ) =\gamma_1$$ but I am stuck with the reminder term.

Side remark: the problem originates in physics, see my older question. In particular, I was able to show that $\langle x \rangle \approx -0.251022$ defined in that question actually equals exactly $-(1+\gamma_0)/(2 \pi)$ where $\gamma_0$ is Euler's constant. This answer I seek here is the only piece missing on my path to closed form $\langle x^2 \rangle$.

• this might be useful: math.stackexchange.com/questions/882621/… nice question btw. (+1) – tired Jan 6 '17 at 10:11
• @tired thanks! Yes, already noticed from SE suggestions, spent a couple of hours with that. Summation has to go to finite $k$, otherwise the corresponding $(u,v)$ integral diverges (as the series diverge). Need to figure out how to deal with divergent terms in that approach... – Slava Kashcheyevs Jan 6 '17 at 10:20

Hint. One may obtain, for $N\ge1$,

\begin{align} \sum_{n=1}^{N} \psi(n) \, \psi'(n)&=\left(N\psi(N)-N+\frac12\right)\psi'(N)+\frac{\psi(N)^2}2- \psi(N)+\frac{\pi^2}{12}-\gamma-\frac{\gamma^2}2 \tag1 \end{align}

equivalently, for $N\ge1$, \small{\begin{align} &\sum_{n=1}^{N} \left(H_{n-1}-\gamma\right)\left(\frac{\pi^2}{6}-H_{n-1,2}\right) \\&=\left(NH_{N-1}-(\gamma+1)N+\frac12\right)\left(\frac{\pi^2}{6}-H_{N-1,2}\right)+\frac{\left(H_{N-1}-\gamma\right)^2}2-\frac{H_{N-1,2}}2-H_{N-1}+\frac{\pi^2}{6}-\frac{\gamma^2}2, \end{align}} then, one may recall the standard asymptotic expansions of the polygamma functions $\psi=\Gamma'/\Gamma$ and $\psi'$, as $X \to \infty$, \begin{align} \psi(X)&= \ln X- \frac{1}{2X}-\frac{1}{12X^2}+\mathcal{O}\left(\frac{1}{X^4}\right) \\\psi'(X)&= \frac{1}{X}+\frac{1}{2 X^2}+\frac{1}{6 X^3}+\mathcal{O}\left(\frac{1}{X^4}\right) \end{align} yielding, as $N \to \infty$,

$$-\frac{(\ln N)^2}2 +\sum_{n=1}^{N} \psi(n) \, \psi'(n)=\color{#9E15E3}{\frac{\pi^2}{12}-\gamma-\frac{\gamma^2}2-1}+\frac{\ln N}{12N^2}-\frac{1}{24N^2}+\mathcal{O}\left(\frac{\ln N}{N^4}\right) \tag2$$

then one gets the desired limit.

To prove $(1)$ one may use a summation by parts with $$f_n=\psi(n),\quad g_{n+1}-g_n=\psi'(n),\quad g_1=\frac{\pi^2}6,\quad n\ge1.$$ The above asymptotic expansion can be obtained at any order.

• According to Wolfy, the value of the constant term is 0.921337593381... which agrees nicely with Claude Leibovici's computation. – marty cohen Jan 8 '17 at 18:25
• Not surprizing at all, one more elegant answer ! Cheers. – Claude Leibovici Jan 9 '17 at 4:34
• What is surprizing is that ISC did not identify the number. – Claude Leibovici Jan 9 '17 at 6:17
• @ClaudeLeibovici Thank you. I've already noticed that ISC does not recognize some numbers involving some standard constants... – Olivier Oloa Jan 9 '17 at 6:33
• that easy...(+1) – tired Jan 9 '17 at 10:04

Considering $$I_2(k)=\sum_{n=1}^k \psi ^{(0)}(n)\,\, \psi ^{(1)}(n)-\frac{\log ^2(k)}{2}$$ being stuck, I generated a few values $$\left( \begin{array}{cc} k & I_2(k) \\ 1000 & -0.92133705940182217478 \\ 2000 & -0.92133744544582151652 \\ 3000 & -0.92133752387787776354 \\ 4000 & -0.92133755278727115778 \\ 5000 & -0.92133756665730197883 \\ 6000 & -0.92133757440092091621 \\ 7000 & -0.92133757917436506891 \\ 8000 & -0.92133758233024150188 \\ 9000 & -0.92133758452841907625 \\ 10000 & -0.92133758612266202362 \end{array} \right)$$ The inverse symbolic calculators I used did not identify the result.
• Very nice! Looks like at least $k^{-1}$ convergence, perhaps even $k^{-2}$. I played my CAS (Mathematica, NSum), reasonably confident in the following digits $I_2 = -0.921337593 \ldots$. – Slava Kashcheyevs Jan 6 '17 at 11:00