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Working on Harmonic numbers, I found this very interesting recurrence relation : $$ H_n = \frac{n+1}{n-1} \sum_{k=1}^{n-1}\left(\frac{2}{k+1}-\frac{1}{1+n-k}\right)H_k ,\quad \forall\ n\in\mathbb{N},n>1$$ My proof of this is quite long and complicated, so I was wondering if someone knows an elegant or concise one. Any idea would be appreciated.

Alternatively, if someone knows a reference that talks about this kind of relation, it would be of great interest for me.


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An equivalent statement of your identity: $$\sum_{k=1}^n\frac{H_k}{n-k+1}=2\sum_{k=1}^n\frac{H_k}{k+1}$$ – J. M. Jul 30 '12 at 3:53
up vote 7 down vote accepted

It helps to visualize the terms in a square of products $1/(ij)$ with $i$ and $j$ running for $1$ to $n$. The sum over the first term contains all products with $i\ne j$ exactly once, whereas the sum over the second term roughly corresponds to the upper left half of the square, but with the left-most column, which adds up to $H_n$, excluded. Thus we have

$$ \def\sub#1{{\scriptstyle{i\ne j}\atop{\scriptstyle i,j\le #1}}} \sum_{k=1}^{n-1}\frac{2}{k+1}H_k=\sum_{\sub n}\frac1{ij}$$


$$-\sum_{k=1}^{n-1}\frac{1}{1+n-k}=H_n-\sum_{i+j\le n+1}\frac1{ij}\;.$$

Substituting this into your equation, multiplying through by $n-1$ and simplifying leads to

$$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=\frac2{n+1}H_n\;,$$

$$\sum_{i+j\le n+1}\frac1{ij}-\sum_{\sub n}\frac1{ij}=2\sum_i\frac1{n+1}\frac1i\;,$$

$$\sum_{i+j\le n+1}\frac1{ij}=\sum_{\sub{n+1}}\frac1{ij}\;.$$

This we can prove by induction: The equation is satisfied for $n=0$, and going from $n$ to $n+1$ adds


to the left-hand side and also


to the right-hand side.

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Thanks a lot. I was looking for this kind of proof at first, but did not find a way to get something as simple as your 5th equation. Great ! – M. M. Jul 30 '12 at 12:23
@M.Mayrand: You're welcome! – joriki Jul 30 '12 at 13:04

Here's a generating function route: as I already mentioned in the comments,

$$\begin{align*} H_n&=\frac{n+1}{n-1}\sum_{k=1}^{n-1}\left(\frac{2}{k+1}-\frac{1}{n-k+1}\right)H_k\\ H_n&=\frac{n+1}{n-1}\left(2\sum_{k=1}^{n-1}\frac{H_k}{k+1}-\sum_{k=1}^{n-1}\frac{H_k}{n-k+1}\right)\\ H_n&=\frac{n+1}{n-1}\left(2\sum_{k=1}^{n}\frac{H_k}{k+1}-\sum_{k=1}^{n}\frac{H_k}{n-k+1}+H_n-\frac{2H_n}{n+1}\right)\\ H_n&=\frac{n+1}{n-1}\left(2\sum_{k=1}^{n}\frac{H_k}{k+1}-\sum_{k=1}^{n}\frac{H_k}{n-k+1}\right)+\frac{n+1}{n-1}H_n-\frac2{n-1}H_n\\ H_n&=\frac{n+1}{n-1}\left(2\sum_{k=1}^{n}\frac{H_k}{k+1}-\sum_{k=1}^{n}\frac{H_k}{n-k+1}\right)+H_n\\ \sum_{k=1}^{n}\frac{H_k}{n-k+1}&=2\sum_{k=1}^{n}\frac{H_k}{k+1} \end{align*}$$

We note that the sum $\sum\limits_{k=1}^{n}\frac{H_k}{n-k+1}$ is in the form of a convolution; thus, its generating function is

$$\left(\sum_{j=1}^\infty H_j x^j\right)\left(\sum_{j=1}^\infty \frac{x^j}{j}\right)=\frac{(\log(1-x))^2}{1-x}$$

The remaining task is to prove that the generating function given above is also the generating function of $2\sum\limits_{k=1}^{n}\frac{H_k}{k+1}$; to that effect, there is the identity


where $H_n^{(k)}=\sum\limits_{j=1}^n \frac1{j^k}$ is a generalized harmonic number. From here and here (see formula 36), we have the generating functions

$$\begin{align*} \sum_{j=1}^\infty H_{j+1}^{(2)}x^{j+1}&=\frac{\mathrm{Li}_2(x)}{1-x}-x\\ \sum_{j=1}^\infty (H_{j+1})^2 x^{j+1}&=\frac{(\log(1-x))^2+\mathrm{Li}_2(x)}{1-x}-x \end{align*}$$

where $\mathrm{Li}_2(x)=-\int_0^x \frac{\log(1-u)}{u}\mathrm du$ is a dilogarithm.


$$\sum_{j=1}^\infty (H_{j+1})^2 x^{j+1}-\sum_{j=1}^\infty H_{j+1}^{(2)}x^{j+1}=\frac{(\log(1-x))^2}{1-x}$$

and we're golden.

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Great proof J.M. Thanks a lot. I continued working on this problem and find another proof. I will post it later today. – M. M. Jul 30 '12 at 12:33

Thanks for your help. Here is another proof that I found today :

From the well known $\zeta(3)=\frac{1}{2}\sum_{k=1}^{\infty}\frac{H_k}{k^2}$, we find $$ \zeta(3)=\sum_{k=1}^{\infty}\frac{H_k}{(k+1)^2} \quad\quad(1) $$ Were $\zeta(z)$ is the Riemann zeta function. But also, $$ \zeta(3)=\frac{1}{2}\int_{0}^{\infty}\frac{t^2}{e^t-1}dt=4\int_{0}^{\pi/2}\tan{x}(\ln{\sin{x}})^2dx=4\int_{0}^{\pi/2}\tan{x}\left(\sum_{k=1}^{\infty}\frac{\cos^{2k}{x}}{2k}\right)^2dx $$ Where I used the substitution $e^{-t}=\sin^2{x}$. By rearranging and using the formula for raising power series to powers (e.g. 0.314 p.17 in Gradshteyn and Ryzhik's Table of Integrals, Series, and Products), $$ \zeta(3)=\sum_{k=0}^{\infty}c_k \int_{0}^{\pi/2}\sin{x}\ \cos^{2k+3}{x}dx=\sum_{k=0}^{\infty}c_k \frac{1}{2}B(1,k+2)=\sum_{k=1}^{\infty}\frac{c_{k-1}}{2(k+1)} $$ $$ \text{where,}\quad c_0=1,\quad c_k=\frac{1}{k}\sum_{i=1}^{k}\frac{3i-k}{i+1}c_{k-i} $$ Equating with (1) and rearranging, we get desired relation.

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Wow, that's quite some detour :-) – joriki Jul 30 '12 at 14:41

This page helped me prove the following related result, so I thought I'd share.

To begin,

$$ \sum_{i=1}^{n}\left ( \frac{1}{x_i} + \frac{1}{x_{n+1-i}}\right ) = 2 \sum_{i=1}^{n}\left ( \frac{1}{x_i} \right ) $$

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

Now, let $X$ be $n$ linear spaced numbers between $a$ and $b$ (and including them).

Then, since in that case $x_i+x_{n+1-i}=a+b$, we have

$$ \left ( a+b \right )\sum_{i=1}^{n}\left ( \frac{1}{x_i x_{n+1-i}} \right ) = 2 \sum_{i=1}^{n}\left ( \frac{1}{x_i} \right ) $$

In other words,

$$ \frac{\sum_{i=1}^{n}\left ( \frac{1}{x_i} \right )}{\sum_{i=1}^{n}\left ( \frac{1}{x_i x_{n+1-i}} \right )} = \frac{a+b}{2} = mean(X) $$

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