# Question Relating Gamma Function to Riemann Zeta function evaluated at integers

I was just reading a paper of Ramanujan entitled "On question 330 of Professor Sanjana" when I got confused regarding a proposition which I am unable to answer. The proposition is:

$\displaystyle f(p) = \frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{ \Bigl[ \Gamma(\frac{P+2}{2}) \Bigr]^{2}}$, then $$\log{f(p)}= \log(\frac{\pi}{2}) - p\log{2} + \frac{p^2}{2} \Bigl(1-\frac{1}{2}\Bigr)S_{2} - \frac{p^3}{3} \Bigl(1-\frac{1}{2^2}\Bigr)S_{3} + \cdots$$

where $\displaystyle S_{n} = \frac{1}{1^n} + \frac{1}{2^n} + \cdots \text{ad inf.}$

Could anyone explain me as to how can i deduce this.

• $S_n$ is called the (Riemann) zeta function. – kennytm Aug 12 '10 at 18:33
• @Kenny TM: Yes, it didn't strike my mind. – anonymous Aug 12 '10 at 18:43

I presume this is a Maclaurin series. I don't know how it comes about, but my instinct would be to expand out the infinite product for $f(p)$ (which becomes an infinite sum for $\log f(p)$) in powers of $p$.
Added (13/8/2010) As a hint: recall that $$\Gamma(p+1)=p\Gamma(p) =e^{-\gamma p}\prod_{k=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}$$ by the infinite product for the gamma function. Applying this to $\Gamma(p+1)/\Gamma(p/2+1)^2$ one gets some nice cancellation. Then take logs and apply the series for $\log(1+x)$. You get a double sum; change the order of summation and hope that the inner sum comes to something like a zeta series.
• Carr's Synopsis 2295 is here. It states that $\log \Gamma(1+n) = (\log\mu - S_1)n + \frac12S_2n^2 - \frac13S_3n^3 + \dots$. Its notation is that µ is "an indefinitely great integer", and it's too late at night for me to parse what that means, but there's a sketch of the proof at the end of the page. – ShreevatsaR Aug 13 '10 at 7:24