I ran across a series and got to wondering how this is so.
We are all familiar with the famous $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}$
But, how can we show:
$\displaystyle\sum_{k=1}^{n}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}-\sum_{k=1}^{n}\frac{\beta(k,n+1)}{k}$
where $\beta$ is the beta function.
Apparently, \displaystyle\sum_{k=1}^{n}\frac{\beta(k,n+1)}{k}={\psi}^{'}(n+1) somehow.
But, the above beta series can be written $\displaystyle\sum_{k=1}^{n}\frac{1}{k}\int_{0}^{1}x^{k-1}(1-x)^{n}dx$.
Also, {\psi}^{'}(n+1)=\displaystyle\sum_{k=0}^{\infty}\frac{1}{(n+k+1)^{2}}
I know that ${\psi}(x)=\int_{0}^{1}\frac{t^{n-1}-1}{t-1}dt-\gamma$
Maybe differentiate w.r.t n and get $\int_{0}^{1}\frac{t^{n-1}ln(t)}{t-1}dt$
Is this related to the incomplete beta function?.
How can we equate these formula, or otherwise, and prove the partial sum?.
There are so many identities involved with Beta, Psi, etc., I get bogged down in all of them. I played around with various things, but have not really gotten anywhere.
Thanks very much.