Series involving harmonic numbers Denote by $H_i$ the $i$-th harmonic number. I conjecture that
$$\lim_{n\ \to\ \infty}\left(\, H_{n}^{2}
-2\sum_{i\ =\ 1}^{n}{H_{i} \over i}\,\right)$$
exists. I have no proof for this. I only have a
vague argument. If you take $H_n \approx \ln n$ then
$$H_n^2 - 2\sum\limits_{i=1}^n \frac{H_i}{i} \approx \ln^2 n - 2 \int\limits_1^n \frac{\ln x}{x} dx = \ln^2 n - 2\left[\frac{\ln^2 x}{2} \right]^n_1 = -\ln^2 1$$
This is far away from being a proof. Question: Does the sequence exist and if so what is its value.
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\begin{align}
\color{#c00000}{\large\sum_{i\ =\ 1}^{n}{H_{i} \over i}}&
=\sum_{i\ =\ 1}^{n}{1 \over i}\sum_{k\ =\ 1}^{i}{1 \over k}
=\sum_{k\ =\ 1}^{n}{1 \over k}\sum_{i\ =\ k}^{n}{1 \over i}
=\sum_{i\ =\ 1}^{n}{1 \over i}
+\sum_{k\ =\ 2}^{n}{1 \over k}\sum_{i\ =\ k}^{n}{1 \over i}
\\[5mm]&=H_{n} + \sum_{k\ =\ 2}^{n}{H_{n} - H_{k - 1} \over k}
=H_{n} + \sum_{k\ =\ 2}^{n}{H_{n} \over k}
-\sum_{k\ =\ 2}^{n}{H_{k - 1} \over k}
\\[5mm]&=H_{n} + H_{n}\pars{H_{n} - 1} - \sum_{k\ =\ 1}^{n}{H_{k} - 1/k\over k}
\\[5mm]&=H_{n}^{2} - \color{#c00000}{\large\sum_{k\ =\ 1}^{n}{H_{k}\over k}}
+\sum_{k\ =\ 1}^{n}{1 \over k^{2}}\quad\imp\quad\boxed{\ds{\quad%
H_{n}^{2} - 2\sum_{i\ =\ 1}^{n}{H_{i} \over i}=-\sum_{k\ =\ 1}^{n}{1 \over k^{2}}}
\quad}
\end{align}

$$\imp\qquad\color{#66f}{\large%
\lim_{n\ \to\infty}\pars{H_{n}^{2} - 2\sum_{i\ =\ 1}^{n}{H_{i} \over i}}}
=-\sum_{k\ =\ 1}^{\infty}{1 \over k^{2}}
=\color{#66f}{\Large -\,{\pi^{2} \over 6\phantom{^{2}}}}\approx {\tt -1.6449}
$$

A: $H_n^2 - 2\sum_{i=1}^n \frac{H_i}{i} = (1 + \frac{1}{2} + \cdots + \frac{1}{n})^2 - 2(\sum_{i=1}^k \sum_{k=1}^n \frac{1}{ik}) = (1 + \frac{1}{2^2} + \cdots + \frac{1}{n^2}) + 2(\sum_{i=k+1}^n \sum_{k=1}^{n-1} \frac{1}{ik}) - 2(\sum_{i=1}^k \sum_{k=1}^n \frac{1}{ik})$
So $\lim_n H_n^2 - 2\sum_{i=1}^n \frac{H_i}{i} = \lim_n (1 + \frac{1}{2^2} + \cdots + \frac{1}{n^2}) = \frac{\pi^2}{6}$
