Question Regarding Integration Within Summation 
It expresses an integral within a summation procedure.

$$\sum _{1}^{\infty}\log (\frac{k+a+1}{k+a})-\log(\frac{1+k+b}{k+b})$$
$$=\sum_1^\infty \log\frac{k+b(k+a+1)}{k+a(k+b+1)}$$
now sum it over 
and expanding the expression gives 
$$=\lim _{k \to \infty}\log \frac{(1+b)(2+a)}{(1+a)(2+b)}\frac {(2+b)(3+a)}{(2+a)(3+b)}.... \frac{(k+a+1)}{(k+b+1)}$$
as the consecutive expressions will cancel out in each iteration
$$=\lim _{k \to \infty} \log\frac{(1+b)(k+a+1)}{(1+a)(k+b+1)}$$
$$=\log\frac{1+b}{1+a}$$
