Need Help Proving An Infinite Series Relating The Gamma Function And Riemann Zeta In a paper entitled 'On Question 330 Of Professor Sanjana' (by Ramanujan), Ramanujan goes on to define a function $f(p)$ as follows
$$f(p) = \frac{1}{1+p} + \frac{1}{2} \frac{1}{3+p} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{5+p} + ...$$
$$= \frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{\Gamma\left(\frac{p+2}{2}\right)^2}$$
And then goes on to say that
$$\ln{f(p)}= \ln\left(\frac{\pi}{2}\right) - p\ln(2)  + \frac{p^2}{2} \Bigl(1-\frac{1}{2}\Bigr)\zeta(2) - \frac{p^3}{3} \Bigl(1-\frac{1}{2^2}\Bigr)\zeta(3) + ...$$
In another post, here they go on to explain that this can be proven using the fact that
$$\Gamma(p+1)=e^{-\gamma p}\prod_{n=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}$$
I tried to do so but got stuck figuiring out what to do next. Here is my attempt:
$$\frac{\Gamma(p+1)}{\Gamma\left(\frac{p+2}{2}\right)^2} = \frac{e^{-\gamma p}\prod_{n=1}^\infty\left(\frac{n}{n+p}\right)e^{p/n}}{\left(e^{-\frac{\gamma p}{2}}\prod_{n=1}^\infty\left(\frac{n}{n+\frac{p}{2}}\right)e^{p/2n}\right)^2}$$
$$= \prod_{n=1}^{\infty}\frac{\left(\frac{n}{n+p}\right)}{\frac{n^2}{\left(n+\frac{p}{2}\right)^2}} = \prod_{n=1}^{\infty}\frac{2n+p}{2n(n+p)}$$
therefore
$$\frac{2^{p+1}}{\pi} f(p) = \prod_{n=1}^{\infty}\frac{2n+p}{2n(n+p)}$$
$$\ln\left(\frac{2^{p+1}}{\pi}\right) + \ln(f(p)) = \sum_{n=1}^{\infty}\ln\left(\frac{2n+p}{2n(n+p)}\right)$$
$$ = \sum_{n=1}^{\infty}\ln\left(2n+p\right) - \ln\left(2n^2+np\right)$$
Here is where I'm not sure what to do next. I know I need to us the series expansion for $\ln(1 \pm x)$, so what do I do? I would appreciate hints and/or solutions. Thanks!
 A: If another approach is allowed, I would propose the following.  First, we note that
$$S(p)=\frac{1}{1+p} + \frac{1}{2} \frac{1}{3+p} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{5+p} + ...=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k-1)!!}{(2k)!!}$$
$$=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k-1)!!(2k)!!}{\big((2k)!!\big)^2}=\sum_{k=0}^\infty\frac{1}{2k+1+p}\frac{(2k)!}{2^{2k}(k!)^2}$$
Next, we use the fact that $\int_0^{2\pi}e^{ikx}=2\pi$ only if $k=0$, otherwise - zero. Therefore, opening the brackets,
$$\int_0^{2\pi}\sin^{2k}xdx=\frac{1}{2^{2k}i^{2k}}\int_0^{2\pi}(e^{ix}-e^{-ix})^{2k}dx=\frac{2\pi}{2^{2k}}\binom{2k}{k}=\frac{2\pi}{2^{2k}}\frac{(2k)!}{(k!)^2}$$
and
$$S(p)=\frac{1}{2\pi}\sum_{k=0}^\infty\frac{1}{2k+1+p}\int_0^{2\pi}\sin^{2k}xdx=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{2k+1+p}\int_0^{\pi/2}\sin^{2k}xdx$$
Using $\displaystyle \frac{1}{2k+1+p}=\int_0^1y^{2k+p}dy$ and changing the order of summation and integration
$$S(p)=\frac{2}{\pi}\int_0^1dy\int_0^{\pi/2}dx\sum_{k=0}^\infty\sin^{2k}(x)\,y^{2k+p}=\frac{2}{\pi}\int_0^1y^pdy\int_0^{\pi/2}\frac{dx}{1-y^2\sin^2x}$$
Using the substitution $t=\tan x$
$$=\frac{2}{\pi}\int_0^1y^pdy\int_0^\infty\frac{dt}{(1-y^2)t^2+1}=\int_0^1\frac{y^p}{\sqrt{1-y^2}}dy=\frac{1}{2}\int_0^1\frac{t^\frac{p-1}{2}}{\sqrt{1-t}}dt$$
$$=\frac{1}{2}B\Big(\frac{p+1}{2};\frac{1}{2}\Big)=\frac{\sqrt\pi}{2}\frac{\Gamma\Big(\frac{p+1}{2}\Big)}{\Gamma\Big(\frac{p+2}{2}\Big)}=\frac{\sqrt\pi}{2}\frac{\Gamma\Big(\frac{p+1}{2}\Big)\Gamma\Big(\frac{p+2}{2}\Big)}{\Gamma\Big(\frac{p+2}{2}\Big)^2}$$
Now, using the duplication formula for gamma function $\Gamma(z)\Gamma\big(z+\frac{1}{2}\big)=2^{1-2z}\sqrt\pi\,\Gamma(2z)$ and the main property $\Gamma(z+1)=z\Gamma(z)$
$$S(p)=\frac{\sqrt\pi}{2}\frac{p}{2}2^{1-p}\sqrt\pi\frac{\Gamma(p)}{\Gamma\Big(\frac{p+2}{2}\Big)^2}=\frac{\pi}{2^{p+1}} \frac{\Gamma(p+1)}{\Gamma\left(\frac{p+2}{2}\right)^2}$$
