How to find this limit with following constraints? Question: If a,b are 2 positive , co-prime integers such that $$\lim _{n \rightarrow \infty}(\frac{^{3n}C_n}{^{2n}C_n})^\frac{1}{n}=\frac{a}{b}$$Then a+b??
I tried to break down the limit to: $$\frac{[(3n)(3n-1)\dot{}\dot{}\dot{})^\frac{1}{n}][n(n-1)(n-2)\dot{}\dot{}\dot{})^\frac{1}{n}]}{[2n(2n-1)(2n-2)\dot{}\dot{}\dot{}]^\frac{2}{n}}$$
but  I'm lost ahead of it. Answer given in my text books is 43. Please guide me
 A: You are lucky to have that exponent $1/n$ because it helps to solve the problem without the use of Stirling's approximation. Instead of Stirling's approximation, we use the following standard theorem on sequences:
Theorem: If $\{a_{n}\}$ is a sequence of positive terms and $a_{n+1}/a_{n}\to L$ as $n\to\infty$ then $a_{n}^{1/n}\to L$ as $n\to\infty$.
Let us define the sequences $\{a_{n} \}, \{b_{n} \} $ by $$a_{n} =\frac{{} ^{3n}C_{n}} {{}^{2n}C_{n}} = \frac{(3n)!n!}{(2n)!(2n)!}, b_{n} =a_{n} ^{1/n}$$ and we need to evaluate the limit of $b_{n} $. To do so we evaluate the limit of $a_{n+1}/a_{n}$. Clearly we have
\begin{align}
\frac{a_{n+1}}{a_{n}}&=\frac{(3n+3)!(n+1)!}{(2n+2)!(2n+2)!}\cdot\frac{(2n)!(2n)!}{(3n)!(n)!}\notag\\
&=\frac{(3n+3)(3n+2)(3n+1)(n+1)}{(2n+2)(2n+1)(2n+2)(2n+1)}\notag\\
&\to \frac{3\cdot 3\cdot 3\cdot 1}{2\cdot 2\cdot 2\cdot 2}\notag\\
&=\frac{27}{16}\notag
\end{align}
Thus by the theorem mentioned above $b_{n} $ also tends to $27/16$ so that $a+b=43$.
A: $\left(\dfrac{^{3n}C_n}{^{2n}C_n}\right)^\frac{1}{n} = \left(\dfrac{(3n)!n!}{(2n)!^2}\right)^{1\over n}$
Stirling approximation is $n! \approx C{n^{1/2}}\dfrac{n^n}{e^n}$. Stuffing it in gives that the limit is $\approx \left(\dfrac{(3n)!n!}{(2n)!^2}\right)^{1\over n}\approx \dfrac{(3n)^3n}{(2n)^2(2n)^2} = \dfrac{27}{16}$.
