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 $$\lim_{n\to\infty}n\left(\sum_{i=1}^{n}f(\dfrac{i}{n})-\int_{0}^{1}f(x)dx\right)=\dfrac{f(1)-f(0)}{2}$$
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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: $$\color{#66f}{\large \lim_{n\ \to\ \infty}n^{2}\braces{n\bracks{H_{2n} - H_{n} -\ln\pars{2}}-{1 \over 4}}} =\color{#66f}{\large \infty} $$