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 $\langle x^2 \rangle$. 
 A: 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.
