$\sum^{\infty}_{n=1} \log(\frac{n+a+b}{n+a} \times \frac{n+b}{n+b+1})$ Prove $$\sum^{\infty}_{n=1} \log(\frac{n+a+1}{n+a} \times \frac{n+b}{n+b+1})=\log\frac{1+b}{1+a}$$
Hints/Answers are appreciated

I think you mean $n+a+1$ not $n+a+b$.
If we put aside the issue of convergence, then using the property $\log(A)+\log(B) = \log(AB)$, we have
$$\sum_{n = 1}^{\infty} \log \left( \frac{n+a+1}{n+a} \cdot \frac{n+b}{n+b+1} \right) = \log \left( \prod_{n = 1}^{\infty} \frac{(a+1+n)(b+n)}{(n+a)(n+b+1)} \right).$$
Note that the $n+b+1$ cancels everything from the $b+n$ term except $b+1$ and same with $n+a+1$. The conclusion follows.
