Find the sum: $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$

Question

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 ?

Answer 1

HINT:

Note that if $a\ne b$, then

$$\begin{align} \int_{1/(b+n)}^{1/(a+n)}\frac1{x+1}\,dx&=\log\left(\frac{1+1/(a+n)}{1+1/(b+n)}\right)\\\\ &=\log\left(\frac{a+n+1}{a+n}\frac{b+n}{b+n+1}\right)\\\\ &\ne \log\left(\frac{a+n+1}{b+n+1}\right) \end{align}$$

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}