partial sum of Basel problem related to series involving Beta function 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.
 A: We wish to prove 
$\displaystyle\sum_{k=1}^{n}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}-\sum_{k=1}^{\infty}\frac{\beta(k,n+1)}{k}.$
(Note: The upper limit in the sum is $\infty$ and not $n$ as given in the question.)
Start with the formula:
$\displaystyle\sum_{k=1}^{\infty}\frac{\beta(k,n+1)}{k}=\displaystyle\sum_{k=1}^{\infty}\frac{1}{k}\int_{0}^{1}x^{k-1}(1-x)^{n}dx.$
The trick is to write the $\displaystyle \frac{1}{k} $ factor on the rhs as an integral:
$\displaystyle \frac{1}{k}=\int_{0}^{1} t^{k-1}dt.$
Substituting this in the above gives
$\displaystyle\sum_{k=1}^{\infty}\frac{\beta(k,n+1)}{k}=\displaystyle\sum_{k=1}^{\infty}\int_{0}^{1}\int_{0}^{1}(xt)^{k-1}(1-x)^{n}dxdt.$
Now, assuming we can interchange the order of summation and integration, the rhs becomes
$\displaystyle\int_{0}^{1}\int_{0}^{1}\sum_{k=1}^{\infty}(xt)^{k-1}(1-x)^{n}dxdt$ $\qquad$ (sum the geometric series)
$=\displaystyle\int_{0}^{1}\int_{0}^{1} \frac{(1-x)^{n}}{1-xt}dxdt$ $\qquad$ (now integrate with respect to t)
$=-\displaystyle\int_{0}^{1} (1-x)^{n}\frac{\ln(1-x)}{x}dx$ $\qquad$ (make the change of variable $x\rightarrow 1-x)$
$=\displaystyle\int_{0}^{1} x^{n}\frac{\ln(x)}{x-1}dx$.
This final integral, as you observed in your question, is equal to
${\psi}^{'}(n+1)=\displaystyle\sum_{k=0}^{\infty}\frac{1}{(n+k+1)^{2}}.$
The proof now goes through.
