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 ? 
 A: HINT:
Note that if $a\ne b$, then
$$\begin{align}
\int_{1/(b+n)}^{1/(a+n)}\frac1{x+1}\,dx&=\log\left(\frac{1+1/(a+n)}{1+1/(b+n)}\right)\\\\
&=\log\left(\frac{a+n+1}{a+n}\frac{b+n}{b+n+1}\right)\\\\
&\ne \log\left(\frac{a+n+1}{b+n+1}\right)
\end{align}$$
A: Btw., the series does not only seem divergent but it is divergent indeed as can be seen as follows:
$$\ln\left(\frac{b+n+1}{a+n+1}\right) = \ln (b+n+1) - \ln(a+n+1) \stackrel{a < \xi_n < b}{=}(b-a)\frac{1}{\xi_n+n+1} \geq (b-a)\frac{1}{\lceil b\rceil+n+1}$$
Hence, the given sum has a divergent tail of the harmonic series as a minorant.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
&\bbox[10px,#ffd]{\left.\sum_{n = 1}^{N}\ln\pars{b + n + 1 \over a + n + 1}
\,\right\vert_{\ 0\ <\ a\ <\ b}} =
\ln\pars{\prod_{n = 1}^{N}{n + b + 1 \over n + a + 1}} =
\ln\pars{{\bracks{b + 2}^{\overline{N}} \over \bracks{a + 2}^{\overline{N}}}}
\\[5mm] = &\
\ln\pars{\Gamma\pars{b + 2 + N}/\Gamma\pars{b + 2} \over
\Gamma\pars{a + 2 + N}/\Gamma\pars{a + 2}}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
\ln\pars{\Gamma\pars{a + 2} \over \Gamma\pars{b + 2}} +
\ln\pars{\root{2\pi}\pars{N + b + 2}^{N + b + 5/2}\expo{-N - b - 2} \over
\root{2\pi}\pars{N + a + 2}^{N + a + 5/2}\expo{-N - a - 2}}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
\ln\pars{\Gamma\pars{a + 2} \over \Gamma\pars{b + 2}} +
\ln\pars{{N^{N + b + 5/2}\,\bracks{1 + \pars{b + 2}/N}^{N} \over
N^{N + a + 5/2}\,\bracks{1 + \pars{a + 2}/N}^{N}}\,\expo{-\bracks{b - a}}}
\\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
\ln\pars{\Gamma\pars{a + 2} \over \Gamma\pars{b + 2}} + \pars{b - a}\ln\pars{N}
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\LARGE\to}\,\,\,\bbx{+\infty}
\end{align}
