# 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.

• Yes, it does exist, and its value is about $-1.6$. – Lucian Feb 14 '14 at 18:37
• A similar limit is $$lim_{n->\infty} \left(2H_n-H_{n^2}\right)=\gamma$$ – Jaume Oliver Lafont Jan 9 '16 at 22:41
• (set m=n in last formula by Eric Naslund in math.stackexchange.com/a/46718/134791) – Jaume Oliver Lafont Jan 9 '16 at 22:49

$$\imp\qquad\color{#66f}{\large% \lim_{n\ \to\infty}\pars{H_{n}^{2} - 2\sum_{i\ =\ 1}^{n}{H_{i} \over i}}} =-\sum_{k\ =\ 1}^{\infty}{1 \over k^{2}} =\color{#66f}{\Large -\,{\pi^{2} \over 6\phantom{^{2}}}}\approx {\tt -1.6449}$$
$H_n^2 - 2\sum_{i=1}^n \frac{H_i}{i} = (1 + \frac{1}{2} + \cdots + \frac{1}{n})^2 - 2(\sum_{i=1}^k \sum_{k=1}^n \frac{1}{ik}) = (1 + \frac{1}{2^2} + \cdots + \frac{1}{n^2}) + 2(\sum_{i=k+1}^n \sum_{k=1}^{n-1} \frac{1}{ik}) - 2(\sum_{i=1}^k \sum_{k=1}^n \frac{1}{ik})$
So $\lim_n H_n^2 - 2\sum_{i=1}^n \frac{H_i}{i} = \lim_n (1 + \frac{1}{2^2} + \cdots + \frac{1}{n^2}) = \frac{\pi^2}{6}$
• Except that every term is negative and that the limit is $-\pi^2/6$. – Did Feb 14 '14 at 20:30