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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:


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.

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But your identity can not be quite right. The partial harmonic sum gives a rational for fixed $n$ and so does the sum of $\beta$ functions. The $\pi^2/6$ is not rational on the other hand... –  Sasha Aug 4 '11 at 16:03
Oh, OK. I did not even notice that. :( But, $\sum_{k=1}^{n}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}-{{\psi}^{'}}(n+1)$. I thought perhaps the series involving beta was somehow equivalent to ${\psi}^{'}(n+1)$. Apparently not. Thank you. I found this interesting link: maths.usyd.edu.au/u/daners/publ/abstracts/zeta2/zeta2.pdf –  Cody Aug 4 '11 at 16:37
The expression of $\psi'(n+1)$ as a series that you recall indicates that $\psi'(n+1)$ is the sum of $1/k^2$ on every $k\ge n+1$. Hence the result you are interested in is simply the decomposition of the total sum into sums over $k\le n$ and over $k\ge n+1$. –  Did Aug 4 '11 at 17:03
According to Maple, $$\eqalign{&\sum_{k=1}^n \frac{\beta(k, n+1)}{k} = \cr &3/2+1/16\,{{}_4F_3(2,2,2,2;\,5/2,3,3;\,1/4)}-3/4\,{\frac {\sqrt {\pi } \Gamma \left( n \right) {n}^{2} {{}_5F_4(1,n+1,n+1,n+1,n+1;\,n,2+n,2+n,n+3/2;\,1/4)}}{ \left( n+1 \right) ^{2}\Gamma \left( n+3/2 \right) {4}^{n}}}\cr &+1/16\, {{}_4F_3(1,2,2,2;\,5/2,3,3;\,1/4)}\cr &-3/4\,{\frac {\sqrt {\pi }\Gamma \left( n \right) n{{}_4F_3(1,n+1,n+1,n+1;\,2+n,2+n,n+3/2;\,1/4)}}{ \left( n+1 \right) ^{2}\Gamma \left( n+3/2 \right) {4}^{n}}}+\Psi \left( 1,n+1 \right) -1/6\,{\pi }^{2}\cr}$$ –  Robert Israel Aug 4 '11 at 17:17
Thanks Robert. I have maple as well and noticed the long, convoluted solution it gave. –  Cody Aug 4 '11 at 20:05
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1 Answer

up vote 4 down vote accepted

We wish to prove


(Note: The upper limit in the sum is $\infty$ and not $n$ as given in the question.)

Start with the formula:


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


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


The proof now goes through.

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Thank you much Peter. I was thinking along the correct lines but didn't, or couldn't, finish. I did not think about writing the 1/k as an integral. –  Cody Aug 4 '11 at 20:03
If I may add something for those interested. I found another interesting closed form for the partial sum of the Basel problem. Let $A_{n}=\int_{0}^{\frac{\pi}{2}}cos^{2n}(x)dx, \;\ B_{n}=\int_{0}^{\frac{\pi}{2}}x^{2}cos^{2n}(x)dx$. Then,$\displaystyle \sum_{k=1}^{n}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}-2\frac{B_{n}}{A_{n}}$ –  Cody Aug 4 '11 at 20:58
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