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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:

$\displaystyle\sum_{k=1}^{n}\frac{1}{k^{2}}=\frac{{\pi}^{2}}{6}-\sum_{k=1}^{n}\frac{\beta(k,n+1)}{k}$

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.

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    Thanks Robert. I have maple as well and noticed the long, convoluted solution it gave.2011-08-04

1 Answers 1

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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.

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    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}}$2011-08-04