A recurrence relation for the Harmonic numbers of the form $H_n = \sum\limits_{k=1}^{n-1}f(k,n)H_k$ 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.
Thanks.
 A: 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
$$
\def\sub#1{{\scriptstyle{i\ne j}\atop{\scriptstyle i,j\le #1}}}
\sum_{k=1}^{n-1}\frac{2}{k+1}H_k=\sum_{\sub n}\frac1{ij}$$
and
$$-\sum_{k=1}^{n-1}\frac{1}{1+n-k}=H_n-\sum_{i+j\le n+1}\frac1{ij}\;.$$
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
$$\sum_{i+j=n+1}\frac1{ij}=\sum_{i=1}^n\frac1i\frac1{n+1-i}$$
to the left-hand side and also
$$2\frac1{n+1}\sum_{i=1}^n\frac1i=\frac1{n+1}\sum_{i=1}^n\left(\frac1i+\frac1{n+1-i}\right)=\sum_{i=1}^n\frac1i\frac1{n+1-i}$$
to the right-hand side.
A: 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
$$2\sum_{k=1}^{n}\frac{H_k}{k+1}=(H_{n+1})^2-H_{n+1}^{(2)}$$
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.
Thus,
$$\sum_{j=1}^\infty (H_{j+1})^2 x^{j+1}-\sum_{j=1}^\infty H_{j+1}^{(2)}x^{j+1}=\frac{(\log(1-x))^2}{1-x}$$
and we're golden.
A: 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.
A: 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,
$$
 \frac{\sum_{i=1}^{n}\left ( \frac{1}{x_i} \right )}{\sum_{i=1}^{n}\left ( \frac{1}{x_i x_{n+1-i}} \right )} = \frac{a+b}{2} = mean(X)
$$
