Find this limits $\lim_{n\to\infty}n^2\bigl(n(H_{2n}-H_{n}-\ln{2})+\frac{1}{4}\bigr)$ Question1:

Find this limits
$$\lim_{n\to\infty}n^2\left(n(H_{2n}-H_{n}-\ln{2})+\dfrac{1}{4}\right)$$
where $$H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}$$

Question 2:

Can we obtain a higher asymptotic expansion?

I know $$ \lim_{n\to\infty}n(H_{2n}-H_{n}-\ln{2})=-\dfrac{1}{4}$$
this following well know
$$\lim_{n\to\infty}n\left(\sum_{i=1}^{n}f(\dfrac{i}{n})-\int_{0}^{1}f(x)dx\right)=\dfrac{f(1)-f(0)}{2}$$
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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It would be easier to use the
Harmonic Number Digamma Representation
$\ds{H_{a} - H_{b}=\Psi\pars{a + 1} - \Psi\pars{b + 1}}$ such that:
\begin{align}&\color{#66f}{\large\lim_{n\ \to\ \infty}
n^{2}\braces{n\bracks{H_{2n} - H_{n} -\ln\pars{2}} + {1 \over 4}}}
\\[5mm]&=\lim_{n\ \to\ \infty}n^{2}\braces{
n\bracks{\dsc{\Psi\pars{2n + 1} - \Psi\pars{n + 1}} - \ln\pars{2}}+{1 \over 4}}
\tag{1}
\end{align}
With the
Digamma Asymptotic Expansion and Recurrence Formula $\ds{\pars{~\mbox{when}\ n \gg 1~}}$:
\begin{align}
&\dsc{\Psi\pars{2n + 1} - \Psi\pars{n + 1}}
=\bracks{\Psi\pars{2n} + {1 \over 2n}} - \bracks{\Psi\pars{n} + {1 \over n}}
\\[5mm]&\sim-\,{1 \over 2n}
+\bracks{\ln\pars{2n} - {1 \over 4n} - {1 \over 48n^{2}}}
-\bracks{\ln\pars{n} - {1 \over 2n} - {1 \over 12n^{2}}}
\\[5mm]&=\ln\pars{2} - {1 \over 4n} + {1 \over 16n^{2}}
\end{align}
The next terms are, at least, of order $\ds{1 \over n^{4}}$. It is clear that:
$$\color{#66f}{\large
\lim_{n\ \to\ \infty}n^{2}\braces{n\bracks{H_{2n} - H_{n} -\ln\pars{2}}-{1 \over 4}}}
=\color{#66f}{\large \infty}
$$
