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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.

Felix Marin
  • 89,464
Volker Turau
  • 161

2 Answers2

4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

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

Felix Marin
  • 89,464
2

$H_n^2 - 2\sum_{i=1}^n \frac{H_i}{i} = (1 + \frac{1}{2} + \cdots + \frac{1}{n})^2 - 2(\sum_{i=1}^k \sum_{k=1}^n \frac{1}{ik}) = (1 + \frac{1}{2^2} + \cdots + \frac{1}{n^2}) + 2(\sum_{i=k+1}^n \sum_{k=1}^{n-1} \frac{1}{ik}) - 2(\sum_{i=1}^k \sum_{k=1}^n \frac{1}{ik})$

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

Marc Romaní
  • 409