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This is an extension of the simpler question [1]

This time we compare sum and integral over the squares of the harmonic numbers (see [1] for definitions)

The sum is

$$f_{s2}(n) = \sum_{k=0}^n H_k^2$$

It can be calculated to give (see (4) in [2])

$$f_{s2}(n) = (n+1) H_n^2 - (2n+1) H_n + 2n$$

Now define the integral

$$f_{i2}(n) = \int_0^n H_x^2 \, dx$$

We are interested in the difference

$$d_{f2}(n) = f_{s2}(n) - f_{i2}(n)$$

The task is to determine the asyptotic behaviour of this differences as $n\to \infty$.

While the asymptotics of $f_{s2}(n)$ is trivially deduced from that of $H_n$ the integral form seems to be tough (compare [3]).

References
[1] Comparison of sum and integral over Harmonic number.
[2] Sum of powers of Harmonic Numbers
[3] Get a good approximation of $\int_0^1 \left(H_x\right)^2 dx$, where $H_x$ is the generalized harmonic number

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  • $\begingroup$ @ Jean Marie Merci. Corrected. $\endgroup$ – Dr. Wolfgang Hintze Oct 3 '17 at 16:59
  • $\begingroup$ By the Euler-Maclaurin summation formula,$$d_{f2}(n)=C+\frac12H_n^2+\frac1{12}\psi^{(0)}(n)+\mathcal O(\psi^{(2)}(n))$$where $C$ is some constant... $\endgroup$ – Simply Beautiful Art Oct 3 '17 at 17:10
  • $\begingroup$ @Simply Beautiful Art : What is the value of the constant C? $\endgroup$ – Dr. Wolfgang Hintze Oct 6 '17 at 11:22
  • $\begingroup$ Some constant C that I haven't figured out. $\endgroup$ – Simply Beautiful Art Oct 6 '17 at 12:30
  • $\begingroup$ @Simply Beautiful Art : May I kindly ask you to explain in more detail how you arrived at your interesting formula, and what the value of C could be, maybe as an answer? $\endgroup$ – Dr. Wolfgang Hintze Oct 6 '17 at 14:01
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Beware: work-in-progress.

Since $H_x=\psi(x+1)+\gamma$ and $\int_{0}^{n}\psi(x+1)\,dx=\log\Gamma(n+1)$, the given problem boils down to computing/approximating $\int_{0}^{n}\psi(x+1)^2\,dx$ (where the basic case $n=1$ is already non-trivial) then applying Stirling's approximation.

$$ H_x = \sum_{a\geq 1}\frac{x}{a(a+x)}, \qquad \int_{0}^{n}H_x^2\,dx=\int_{0}^{n}\sum_{a,b\geq 1}\frac{x^2}{ab(a+x)(b+x)}\,dx $$ lead to: $$\begin{eqnarray*} \int_{0}^{n}H_x^2\,dx &=& \sum_{c\geq 1}\int_{0}^{n}\frac{x^2\,dx}{c^2(c+x)^2}+\sum_{\substack{a,b\geq 1\\ a\neq b}}\frac{1}{ab(a-b)}\int_{0}^{n}\left(\frac{ax}{a+x}-\frac{bx}{b+x}\right)\,dx\\&=&S_1(n)+2\sum_{a>b\geq 1}\frac{1}{ab(a-b)}\int_{0}^{n}\left(\frac{ax}{a+x}-\frac{bx}{b+x}\right)\,dx\\&=&S_1(n)+2\,S_2(n)\end{eqnarray*}$$ where: $$\begin{eqnarray*} S_1(n) &=& n\,\zeta(2)+H_n+2\sum_{a\geq 1}\frac{\log(a)-\log(a+n)}{a}\\ &=& n\,\zeta(2)+H_n+2\int_{0}^{+\infty}(1-e^{-nt})\log(1-e^{-t})\frac{dt}{t}\\&=&n\,\zeta(2)+H_n+2\color{blue}{\int_{0}^{1}\frac{(1-u^n)\log(1-u)}{u\log u}\,du} \end{eqnarray*}$$ and $$ S_2(n) = \sum_{b\geq 1}\sum_{s\geq 1}\frac{1}{bs(b+s)}\int_{0}^{n}\frac{sx^2\,dx}{(b+x)(b+s+x)}\,dx. $$

It follows that the first terms of the wanted asymptotic expansion are not that difficult to find, but the constant term depends on not-so-common integrals like the blue one, which we already met when trying to find a closed form for $\int_{0}^{1}H_x^2\,dx$.

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