Let $\displaystyle{\displaylines{a,b \in \mathbb{R}_+^*}}$ and consider the sequence :
$\displaystyle{\displaylines{u_n = \prod_{k=0}^{n} \frac{a+k}{b+k}}}$
Prove with 2 different methods that :
$\displaystyle{\displaylines{\sum u_n \text{ converge } \iff b-a > 1}}$
First Method:
Consider Euler Gamma function :
$\displaystyle{\displaylines{\forall x \in \mathbb{R}_{+}^{*} \ : \ \Gamma(x)=\lim_{n \to +\infty} \frac{n^x}{x} \prod_{k=1}^{n}\left( 1 + \dfrac{x}{k} \right)^{-1}}}$
We have :
$\displaystyle{\displaylines{\begin{array}{rcl}u_n & = & \displaystyle \prod_{k=0}^{n} \frac{a+k}{b+k} \\ & = & \displaystyle \dfrac{a}{b} \prod_{k=1}^{n}\left(1 + \dfrac{a}{k} \right) \prod_{k=1}^{n}\left( 1 + \dfrac{b}{k} \right)^{-1} \\ & \underset{n \to +\infty}{\sim} & \dfrac{n^a}{\Gamma(a)} \Gamma(b) n^{-b} \\ & \underset{n \to +\infty}{\sim} & \dfrac{\Gamma(b)}{\Gamma(a)} \dfrac{1}{n^{b-a}}\end{array}}}$
$\displaystyle{\displaylines{\sum \dfrac{1}{n^{b-a}}}}$ converge iff $\displaystyle{\displaylines{b-a > 1}}$, then we have the result :
$\displaystyle{\displaylines{\sum u_n \text{ converge } \iff b-a > 1}}$
Second Method:
We have :
$\displaystyle{\displaylines{\begin{array}{rcl}\dfrac{u_{n+1}}{u_n} & = & \displaystyle \prod_{k=0}^{n} \left( \frac{b+k}{a+k} \right) \prod_{k=0}^{n+1} \left( \frac{a+k}{b+k} \right) \\ & = & \dfrac{a+n+1}{b+n+1} \\ & = & 1-\dfrac{b-a}{b+n+1} \\ & = & 1 - \dfrac{b-a}{n} \left(1 + o(1) \right) \\ & = & 1 - \dfrac{b-a}{n} + o\left( \dfrac{1}{n} \right)\end{array}}}$
Using Raabe–Duhamel's test :
if $\displaystyle{\displaylines{b-a > 1}}$, then $\displaystyle{\displaylines{\sum u_n}}$ converge.
if $\displaystyle{\displaylines{b-a < 1}}$, then $\displaystyle{\displaylines{\sum u_n}}$ diverge.
if $\displaystyle{\displaylines{b-a=1}}$, then : $\displaystyle{\displaylines{u_n = \frac{a}{a+n+1} \sim \frac{a}{n}}}$ then $\displaystyle{\displaylines{\sum u_n}}$ diverge.
Conclusion :
$\displaystyle{\displaylines{\sum u_n \text{ converge } \iff b-a > 1}}$
