A question related to harmonic numbers

Question

For $n \geq 1$ fixed, I want to know how to compute the double sum

\begin{equation} \sum_{i=1}^{n-1} \sum_{j=i+1}^n \frac{1}{i \cdot j}. \end{equation}

In particular, can we say anything about the order of this sum in terms of n ? Thanks a lot.

Answer 1

if you set $$ s_n= \sum_{k=1}^n \frac1{k^2} \\ h_n= \sum_{k=1}^n \frac1{k} $$ then $$ \sum_{i=1}^{n-1} \sum_{j=i+1}^n \frac{1}{i \cdot j} = \frac12 \left( h_n^2 - s_n \right) $$

Answer 2

First change order of summation:

$$\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{1}{ij}=\sum_{j=1}^{n-1}\sum_{i=1}^{n-j}\frac{1}{i(j+i)}$$

Now for fixed $j$ we can compute $\sum_{i=1}^{n-j}\frac{1}{i(j+i)}$:

$$\sum_{i=1}^{n-j}\frac{1}{i(j+i)}=\sum_{i=1}^{n-j}\frac{1}{j}\left(\frac{1}{i}-\frac{1}{j+i}\right)=\frac{1}{j}\sum_{i=1}^{n-j}\left(\frac{1}{i}-\frac{1}{j+i}\right)=\frac{1}{j}(H_{n-j}-(H_{n}-H_{j-1}))=\frac{1}{j}\left(-H_{n}+H_{n-j}+H_{j-1}\right)$$

So:

$$\sum_{j=1}^{n-1}\sum_{i=1}^{n-j}\frac{1}{i(j+i)}=\sum_{j=1}^{n-1}\frac{1}{j}\left(-H_{n}+H_{n-j}+H_{j-1}\right)=-H_nH_{n-1}+\sum_{j=1}^{n-1}\frac{1}{j}\left(H_{n-j}+H_{j-1}\right)=\\=-\sum_{j=1}^{n-1}\sum_{i=1}^{n-j}\frac{1}{i(j+i)}+\sum_{j=1}^{n-1}\frac{1}{j}\left(\sum_{i=1}^{j}\frac{1}{i}-\frac{1}{j}\right)=\\=-\sum_{j=1}^{n-1}\sum_{i=1}^{n-j}\frac{1}{i(j+i)}+H_n^2-\sum_{j=1}^{n}\frac{1}{j^2}$$

Answer 3

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$\ds{\tt\mbox{Note that}}$:

\begin{align}&\color{#66f}{\large\sum_{i\ =\ 1}^{n - 1}\sum_{j\ =\ i + 1}^{n}% {1 \over i\,j}} =\sum_{i\ =\ 1}^{n - 1}{1 \over i} \pars{\sum_{j\ =\ 1}^{n}{1 \over j} - \sum_{j\ =\ 1}^{i}{1 \over j}} =\sum_{i\ =\ 1}^{n - 1}{1 \over i}\pars{H_{n} - H_{i}} =H_{n - 1}H_{n} - \color{#c00000}{% \sum_{i\ =\ 1}^{n - 1}{H_{i} \over i}}\,\,\,\pars{1} \end{align}

However \begin{align}&\color{#c00000}{\sum_{i\ =\ 1}^{n - 1}{H_{i} \over i}} =\sum_{i\ =\ 1}^{n - 1}{1 \over i}\int_{0}^{1}{1 - t^{i} \over 1 - t}\,\dd t =\sum_{i\ =\ 1}^{n - 1}{1 \over i} \int_{0}^{1}\ln\pars{1 - t}\pars{-i\,t^{i - 1}}\,\dd t \\[5mm]&=-\int_{0}^{1}\ln\pars{1 - t}\sum_{i\ =\ 1}^{n - 1}t^{i - 1}\,\dd t =-\int_{0}^{1}\ln\pars{1 - t}\,{1 - t^{n - 1} \over 1 - t}\,\dd t \\[5mm]&=\int_{t\ =\ 0}^{t\ =\ 1}\pars{1 - t^{n - 1}}\, \dd\bracks{\ln^{2}\pars{1 - t}} =\pars{n - 1}\int_{0}^{1}t^{n - 2}\ln^{2}\pars{1 - t}\,\dd t \\[5mm]&=\pars{n - 1}\lim_{\mu\ \to\ 0}\partiald[2]{}{\mu} \int_{0}^{1}t^{n - 2}\pars{1 - t}^{\mu}\,\dd t =\pars{n - 1}\lim_{\mu\ \to\ 0}\partiald[2]{}{\mu}\bracks{% \Gamma\pars{n - 1}\Gamma\pars{\mu + 1} \over \Gamma\pars{n + \mu}} \\[5mm]&=\pars{n - 1}!\lim_{\mu\ \to\ 0}\partiald[2]{}{\mu}\bracks{% \Gamma\pars{\mu + 1} \over \Gamma\pars{n + \mu}} =\Psi^{2}\pars{n} + 2\gamma\Psi\pars{n} - \Psi'\pars{n} + \zeta\pars{2} + \gamma^{2} \end{align}

Replacing in $\pars{1}$: \begin{align}&\color{#66f}{\large\sum_{i\ =\ 1}^{n - 1}\sum_{j\ =\ i + 1}^{n}% {1 \over i\,j}} =\color{#66f}{\large H_{n - 1}H_{n} -\Psi^{2}\pars{n} - 2\gamma\Psi\pars{n} + \Psi'\pars{n} - \zeta\pars{2} - \gamma^{2}} \end{align}