<s>[INST] Find the sum: $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ Let $0<a<b$, I would like to compute the sum 
$$\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right).$$
But first I am worrying that a test convergence might lead to the divergence of this series
What do I miss here? 
$$\begin{split}\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)&= \sum_{n=1}^{\infty} \int^{\frac1{a+n}}_{\frac1{b+n}}\frac{dx}{x+1}\\
&= \sum_{n=1}^{\infty} \int_{a+n}^{b+n}\frac{dt}{t(t+1)}~~~~(t= 1/x)\\
&= \sum_{n=1}^{\infty} \int_{a}^{b}\frac{dt}{(t+n)(t+n+1)}\\
&=\int_{a}^{b}dt \sum_{n=1}^{\infty} \frac{1}{t+n}-\frac{1}{t+n+1}~~~(\text{Monotone convergence})\\
&= \int_{a}^{b}\frac{dt}{t+1} ~~~~(\text{by Telescoping sum})\\
&= \ln\left(\frac{b+1}{a+1}\right) \end{split}$$
However the series seems $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ not to be convergent.
Have I missed something ? 
 [/INST] HINT:
Note that if $a\ne b$, then
$$\begin{align}
\int_{1/(b+n)}^{1/(a+n)}\frac1{x+1}\,dx&=\log\left(\frac{1+1/(a+n)}{1+1/(b+n)}\right)\\\\
&=\log\left(\frac{a+n+1}{a+n}\frac{b+n}{b+n+1}\right)\\\\
&\ne \log\left(\frac{a+n+1}{b+n+1}\right)
\end{align}$$
</s>