Prove that
$$\prod_{n=1}^\infty{n(n+a+b)\over(n+a)(n+b)} = {\Gamma(a+1)\Gamma(b+1)\over\Gamma(a+b+1)}$$
whenever $a$ and $b$ are positive. Using the product formula for $\sin\pi s$, give another proof that $\Gamma(s)\Gamma(1-s) =\pi/\sin\pi s$.
My attempt: First of all, I guess the assumption $a,b$ are positive can be dropped. I think it should be dropped because we need to use that product formula to show $\Gamma(s)\Gamma(1-s) = \pi/\sin\pi s$.
Recall the product formula for entire function $1/\Gamma(s)$:
$${1\over\Gamma(s)}= e^{\gamma s}s\prod_{n=1}^\infty\left(1+{s\over n}\right)e^{-{s\over n}}$$
for all $s\in\Bbb C$. In particular, term-by-term multiplication of infinite products is allowed.
\begin{align*}
{\Gamma(a+1)\Gamma(b+1)\over\Gamma(a+b+1)} & = {e^{\gamma(a+b+1)}(a+b+1)\prod_{n=1}^\infty\left(1+{a+b+1\over n}\right)e^{-{a+b+1\over n}}\over e^{\gamma(a+1)}(a+1)\prod_{n=1}^\infty\left(1+{a+1\over n}e^{-{a+1\over n}}\right)e^{\gamma(b+1)}(b+1)\prod_{n=1}^\infty\left(1+{b+1\over n}\right)e^{-{b+1\over n}}}\\
& = {e^{\gamma(a+b+1)}(a+b+1)\over e^{\gamma(a+b+2)}(a+1)(b+1)}\prod_{n=1}^\infty{\left(1+{a+b+1\over n}\right)e^{-{a+b+1\over n}}\over \left(1+{a+1\over n}\right)\left(1+{b+1\over n}\right)e^{-{a+b+2\over n}}}\\
& = {e^{\gamma(a+b+1)(a+b+1)}\over e^{\gamma(a+b+2)}(a+1)(b+1)}\prod_{n=1}^\infty{(a+b+n+1)n\over (a+n+1)(b+n+1)}e^{1\over n}\\
& = e^{-\gamma}\prod_{n=1}^\infty{n(a+b+n)\over(a+n)(b+n)}e^{1\over n}\\
\end{align*}
Now observe that using $\Gamma(1) =1$,
\begin{align*}
1 & = e^{\gamma}\prod_{n=1}^\infty\left(1+{1\over n}\right)e^{-{1\over n}}\\
e^{-\gamma} & = \prod_{n=1}^\infty{n+1\over n}e^{-{1\over n}}.\\
\end{align*}
Hence,
\begin{align*}
e^{-\gamma}\prod_{n=1}^\infty{n(a+b+n)\over(a+n)(b+n)}e^{1\over n} & = \prod_{n=1}^\infty{(n+1)(a+b+n)\over(a+n)(b+n)}\\
& = \prod_{n=1}^\infty{n(a+b+n)\over (a+n)(b+n)}\\
\end{align*}
Does it make sense?
Best Answer
It looks correct to me, reading between the lines of the LaTex typos!
You can also do it like this:
Fix a $b$ in the half plane $H:=\{z\in\Bbb C:\Re z>0\}$ and consider the complex function: $$F_b:H\to\Bbb C$$Defined by: $$s\mapsto\sum_{n=1}^\infty[\log(n)+\log(n+s+b)-\log(s+n)-\log(b+n)]$$Where appears the principal logarithm. $F_b$ is holomorphic everywhere on its domain and we can consider: $$\begin{align}F_b'(s)&=\sum_{n=1}^\infty\left[\frac{1}{n+s+b}-\frac{1}{n+s}\right]\\&=\sum_{n=1}^\infty\left[\left(\frac{1}{n}-\frac{1}{n+s}\right)-\left(\frac{1}{n}-\frac{1}{n+s+b}\right)\right]\\&=\left(-\gamma+\sum_{n=1}^\infty\left[\frac{1}{n}-\frac{1}{n+s}\right]\right)-\left(-\gamma+\sum_{n=1}^\infty\left[\frac{1}{n}-\frac{1}{n+s+b}\right]\right)\\&=\psi(s+1)-\psi(s+b+1)\end{align}$$
And we also have $\lim_{s\to 0,s\in H}F_b(s)=0$ hence: $$F_b(s)=\oint_0^s[\psi(z+1)-\psi(z+b+1)]\,\mathrm{d}z=\log\Gamma(s+1)-\log\Gamma(s+b+1)+\log\Gamma(b+1)$$
Exponentiating: $$\exp(F_b(s))=\frac{\Gamma(s+1)\Gamma(b+1)}{\Gamma(s+b+1)}$$For all $s\in H$. But the right hand side is analytic outside of $\Bbb C\setminus(\Bbb Z_{<0}\cup(\Bbb Z_{<0}-b))$, and the left hand side is the product: $$s\mapsto\prod_{n=1}^\infty\frac{n(n+s+b)}{(n+s)(n+b)}$$Which is analytic in the same region. We can then claim a total equality by the identity theorem, and you may specify $s=a$.
I've just noticed you were only supposed to prove it for positive $a,b$, in which case my proof over $H$ works fine without analytic continuation.