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.
 A: 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.
A: Take a look at equation 34 on http://mathworld.wolfram.com/GammaFunction.html After taking the log of both sides, the zeta function appears in essentially the same form as in your equation. They reference a 1968 paper by Wrench for more information. I haven't worked through the details myself, but the similarities are striking, and it looks like a promising approach.
