Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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: – 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
up vote 5 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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.