I have the following recurrence formula: $$ w_i = \frac{n(n-1)}{(n-i)(i+1)} + \frac{i-1}{i+1}w_{i-1} $$ that is defined for $i = 1,\cdots, n$.
I would like to write it in terms of just $i$ and $n$, without the $w_{i-1}, so I imagine the formula would end up being a summation and a product.
I expanded a little bit just to identify some structure in the recurrence \begin{align} w_i =\frac{n(n-1)}{(n-i)(i+1)} + \frac{i-1}{i+1}(\frac{n(n-1)}{(n-(i-1))(i)} + \frac{i-2}{i}w_{i-2}) \\ w_i =\frac{n(n-1)}{(n-i)(i+1)} + \frac{i-1}{i+1}(\frac{n(n-1)}{(n-(i-1))(i)} + \frac{i-2}{i}(\cdots (\cdots (\cdots (\cdots))))) \\ \end{align}
But I'm having a hard time wrapping my head around all the nested parenthesis, and how to write this succinctly with $\sum$ and $\prod$. Could someone help?
Let $w_i=\frac{z_i}{i(i+1)}$ to make $$z_{i}=z_{i-1}+\frac{i (n-1) n}{n-i}$$ which let suspecting harmonic numbers in the solution.
The remaining was done using a CAS which gave $$z_i=C-i(n-1)+n(n-1) \left(H_{-n}-H_{i-n}\right)$$