Weird sum that is almost definitely not $\sqrt 2$

Question

I have not the ability to compute more than four digits of

$$\sum_{n=1}^\infty \frac{1}{n^2 H_n^{(\ln n)}}$$

$H_n^{(m)} = \sum_{k=1}^n \frac{1}{k^m}$ is the generalized harmonic number.

I know this is the weirdest sum and it offers me no actual interest. All I know is that the decimal number starts off as $1.414...$ and I want to settle my mind that it is not actually $\sqrt 2$. That would be crazy. I have no reason to expect it. I just want some confirmation.

My calculations were from Desmos here https://www.desmos.com/calculator/helb1dgf1g.

Answer 1

After interpolation by $H_n^{(s)}=\zeta(s)-\zeta(s,n+1)$ using Hurwitz zeta function, one might apply accelerated summation methods, such as Abel–Plana or (discrete) Euler–Maclaurin.

The latter is implemented in PARI/GP's sumnum command. The computation of $\zeta(s,a)$ using out-of-the-box zetahurwitz runs too slow for large values of $a$. A working solution is the asymptotics $$\zeta(s+1,a)\underset{a\to\infty}{\phantom{\big[}\asymp\phantom{\big]}}\frac1s\sum_{n=0}^{(\infty)}\binom{-s}{n}\frac{B_n}{a^{n+s}}.$$

I'm using $\sum_{n=0}^N$ at $a\geqslant M$, being too lazy to automate the choice of $M$ and $N$:

hzfasymp(s, a, N) = sum(n = 0, N, bernfrac(n) * binomial(-s, n) / a^n) / (s * a^s);
hzftuned(s, a, M, N) = if(a < M, zetahurwitz(s, a), hzfasymp(s - 1, a, N));
experiment(L, M, N) = sum(n = 1, L - 1, 1 / n^2 / sum(k = 1, n, k^(-log(n)))) \
 + sumnum(n = L, 1 / n^2 / (zeta(log(n)) - hzftuned(log(n), n + 1, M, N)));

so that experiment(10, 10^5, 40) produces more than $100$ correct digits, beginning with $$1.41414021781626730415277043584162320768489153366403329952828864498511\dots$$