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Let a,b be positive number then how to find the limit of $x_n=\frac{(a)(a+1)(a+2)..(a+n)}{(b)(b+1)(b+2)..(b+n)}$ when $n\rightarrow \infty $ when a=b obviously the limit is 1 but what about $a<b$ ?

this problem reminded me of the limit $0<A_n=\frac{1*3*5*..*(2n-1)}{2*4*6*..*(2n)} \rightarrow 0 $ I can show this by denote $B_n=\frac{2*4*6*..*(2n)}{3*5*7*..*(2n+1)}$ then $A_n<B_n$ and $A_nB_n=\frac{1}{(2n+1)}$ so $A_n<\sqrt{\frac{1}{2n+1}}\rightarrow0$ when $n\rightarrow \infty$ but the same method seem to be useless in the above problem.I now also consider using delta-epsilon method or even using Gamma function to assure that the limit should be zero.What method should I use?

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    $\begingroup$ $a(a+1)(a+2)..(a+n) = \Gamma (a+n+1) / \Gamma (a)$. Have you tried Stirling's approximation? $\endgroup$ – Stephen Herschkorn Jul 21 '13 at 16:39
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(Since $x_n>x_{n+1}>0$ $\lim_{n\to\infty} x_n$ does exist and $\geq0$.) Applying the inequality $1+x\leq e^x$ (which is valid for any real $x$) we obtain $$ \frac{a+k}{b+k}=\frac{b+k+a-b}{b+k}=1+\frac{a-b}{b+k}\leq e^{\frac{a-b}{b+k}}. $$ Using this inequality we obtain $$ 0\leq x_n=\frac{(a)(a+1)(a+2)..(a+n)}{(b)(b+1)(b+2)..(b+n)}\leq\exp\left( (a-b)\sum_{k=0}^{n}\frac{1}{b+k} \right). $$ Here $a-b<0$ and $\sum_{k=0}^{n}\frac{1}{b+k}$ tends to $\infty$ as $n\to\infty$. So the limit is $0$.

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