How to calculate $\lim_{n\rightarrow\infty}\frac{a(a+1)(a+2)\cdots(a+n)}{b(b+1)(b+2)\cdots(b+n)}$ How to calculate $$\lim_{n\rightarrow\infty}\frac{a(a+1)(a+2)\cdots(a+n)}{b(b+1)(b+2)\cdots(b+n)}$$
where $a>0, b>0?$
I could not calculate the limit with ordinary tool(ratio test, squeeze ...).
Does anyone have an idea of efficiency to find this limit?
Thank you!
 A: Since:
$$\log\left(\frac{a+n}{b+n}\right)=\frac{a-b}{n}+\frac{b^2-a^2}{2n^2}\cdot\left(1+O\left(\frac{1}{n}\right)\right) $$
we have that 
$$\prod_{k=1}^{n}\frac{a+k}{b+k}\sim C\cdot\exp\sum_{k=1}^{n}\frac{a-b}{k}$$
and since:
$$\sum_{k=1}^{n}\frac{1}{k}=H_n = \log n+\gamma+O\left(\frac{1}{n}\right)$$
the limit is $+\infty,1$ or $0$ depending on $a>b,a=b$ or $a<b$.
A: An approach that doesn't use asymptotics:
We consider the case $a>r$, that is, $a = b + r$ with $r>0$.  Our sequence satisfies the recurrence
$$
f(n) = f(n-1)\frac{b+r+n}{b+n} = f(n-1)\left(1 + \frac{r}{b+n} \right)
$$
We note that our sequence is positive and monotonically increasing.  It follows that
$$
f(n) - f(n-1) = f(n-1)\left(1 + \frac{r}{b+n} \right) - f(n-1) =\\
\frac{r}{b + n}f(n-1) >
\frac{r}{b + n}f(0) = \frac{a}{b}\frac{r}{b+n}
$$
We then note that
$$
\lim_{n \to \infty} f(n) = f(0) + 
\lim_{n \to \infty} \sum_{k=1}^n (f(k) - f(k-1)) >
f(0) + f(0)\lim_{n \to \infty} \sum_{k=1}^n \frac{r}{b + k}
$$
By the divergence of the sum, we conclude that the limit is $\infty$.

In the case $a = b$, note that $f(n) = 1$, so our limit is $1$.

In the case $a < b$, our limit can be rewritten as
$$
\lim_{n\rightarrow\infty}\left(\frac{b(b+1)(b+2)\cdots(b+n)}{a(a+1)(a+2)\cdots(a+n)}\right)^{-1}
$$
so that from our previous analysis, the limit is $0$.
A: Another (close though) method is to evaluate the ratio between two consecutive terms. Let $a_n$ be your sequence.
$V_n = \frac{a_{n}}{a_{n-1}} =\frac{a+n}{b+n} = 1 + \frac{a-b}{n} + O(\frac{1}{n^2}) $
$a_n$ >0, we can consider $u_n=\ln(a_n)$:
$ u_{n}-u_{n-1} = \ln(1 + \frac{a-b}{n} + O(\frac{1}{n^2})) = \frac{a-b}{n} + O(\frac{1}{n^2})  $
Summing partial sums of diverging series, you get: 
$ u_n -u_1 = (a-b)*H_n + G_n $ with $H_n$ the partial sum of the $\frac{1}{k}$, and $G_n$ the partial sum of a converging series. 
It is well-known that : $H_n$ ~ ln(n) , hence you get:
$ u_n -u_1 $ ~ $u_n$ ~ $(a-b)\ln(n)$
Ie : $a_n = \exp(u_n)$ ~ $ K*n^{a-b}$
The conclusion are the same: diverges if a > b; =1 if a=b; $\rightarrow 0$ if a < b
One little precision : if b-a > 1, then $\sum a_n$ converges. 
A: Your limit can be rewritten as $$\displaystyle \lim_{n\to +\infty} \frac{\frac{\Gamma(a+n+1)}{\Gamma(a)}}{\frac{\Gamma(b+n+1)}{\Gamma(b)}} =\frac{\Gamma(b)}{\Gamma(a)}\lim_{n\to +\infty}\frac{\Gamma(a+n+1)}{\Gamma(b+n+1)},$$ which using Stirling's approximation (note that if $x$ is a positive real number, it is indeed $\Gamma(x)\sim\sqrt{2\pi (x-1)} (x-1)^{x-1} e^{-x+1}$) is equal to $$\displaystyle \frac{\Gamma(b)}{\Gamma(a)} \lim_{n\to +\infty} \frac{\sqrt{2\pi(a+n)}\left(\frac{a+n}{e}\right)^{a+n}}{\sqrt{2\pi(b+n)}\left(\frac{b+n}{e}\right)^{b+n}}=\frac{\Gamma(b)}{\Gamma(a)}\lim_{n\to +\infty}\frac{\left(\frac{a+n}{e}\right)^{a+n}}{\left(\frac{b+n}{e}\right)^{b+n}}= \\ \frac{\Gamma(b)}{\Gamma(a)}\lim_{n\to +\infty}\frac{\left( a+n\right)^{a+n}e^{b+n}}{\left(b+n \right)^{b+n}e^{a+n}}=\\e^{b-a}\frac{\Gamma(b)}{\Gamma(a)}\lim_{n\to +\infty}\frac{\left( a+n\right)^{a+n}}{\left(b+n \right)^{b+n}}=
\left\{ 
\begin{array}{c}
+\infty &\text{if} \ a>b \\ 
1 &\text{if} \ a=b \\ 
0 &\text{if} \ a<b
\end{array}
\right. 
$$
A: The case $a=b$ is trivial (limit is $1$). Let's focus on $a\neq b$. Take the logarithm:
$$\ln u_n = \sum_{k=0}^n \ln\frac{a+k}{b+k} = \sum_{k=0}^n \ln\frac{1+\frac{a}{k}}{1+\frac{b}{k}}\;.$$
Now, when $k\to\infty$ you have $\ln\frac{1+\frac{a}{k}}{1+\frac{b}{k}}\sim \frac{a-b}{k}$. By theorems of comparisons of series, this implies that
$$
\ln u_n \operatorname*{\sim}_{n\to\infty} (a-b)\ln n\xrightarrow[n\to\infty]{} \begin{cases}
+\infty &\text{ if } a > b \\ 
-\infty &\text{ if } a < b.
\end{cases}
$$
To sum up, this gives you that $u_n\xrightarrow[n\to\infty]{} \begin{cases}
+\infty &\text{ if } a > b \\ 
1 &\text{ if } a = b \\ 
0 &\text{ if } a < b.
\end{cases}$
