Binomial limit

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left [ \frac{\binom{3n}{n}}{\binom{2n}{n}} \right ]^\frac1n$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left [ \frac{\frac{ (3n)!}{(2n)! n!} }{\frac{(2n)!}{n!n!}} \right ]^\frac1n$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left [ \frac {(3n)! n!}{(2n)! (2n)!} \right ] ^\frac1n$

By Stirling's approximation :

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left [ \frac{ \sqrt{2 \pi \cdot 3n} \left ( \frac{3n}{e} \right ) ^{3n} \cdot \sqrt{2 \pi \cdot n} \left ( \frac{n}{e} \right ) ^{n} }{\sqrt{2 \pi \cdot 2n} \left ( \frac{2n}{e} \right ) ^{2n} \cdot \sqrt{2 \pi \cdot 2n} \left ( \frac{2n}{e} \right ) ^{2n} } \right] ^\frac1n$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left [ \frac{\sqrt{3}}{2} \cdot \left ( \frac{27}{16} \right)^n \right ]^\frac1n$

$\color{#20A900} \boxed{ \large \displaystyle L = \frac{27}{16} }$

$\color{#3D99F6} \large \displaystyle a = 27, b = 16, \boxed{ \large \displaystyle a+b = 43}$

$\begin{aligned} \displaystyle L &= \lim_{n \to \infty} \left(\dfrac{\binom{3n}{n} }{\binom{2n}{n} }\right)^{\frac{1}{n}} \\ \displaystyle &= \lim_{n \to \infty} \left( {\frac{(3n)!}{n! (2n)!}}{\huge /}{\frac{(2n)!}{(n!)^2}} \right)^{\frac{1}{n}} \\ \displaystyle &= \lim_{n \to \infty} \left( {\frac{(3n)!}{(2n)!}}{\huge /}{\frac{(2n)!}{(n!)}} \right)^{\frac{1}{n}} \\ \displaystyle &= \lim_{n \to \infty} \left( \dfrac{\displaystyle \prod_{r=1}^n (2n+r)}{\displaystyle \prod_{r=1}^n (n+r)} \right)^{\frac 1n} \\ \displaystyle &= \exp{ \left[ \lim_{n \to \infty} \dfrac 1n \ln \left( \dfrac{\displaystyle \prod_{r=1}^n (2n+r)}{\displaystyle \prod_{r=1}^n (n+r)} \right) \right] } \\ \displaystyle &= \exp{ \left[ \lim_{n \to \infty} \dfrac 1n \ln \left( \prod_{r=1}^n (2n+r) \right) - \dfrac 1n \ln \left( \prod_{r=1}^n (n+r) \right) \right] } \\ \displaystyle &= \exp{ \left[ \lim_{n \to \infty} \dfrac 1n \sum_{r=1}^n \ln (2n+r) - \dfrac 1n \sum_{r=1}^n \ln (n+r) \right] } \\ \displaystyle &= \exp{ \left[ \lim_{n \to \infty} \dfrac 1n \sum_{r=1}^n \ln \left(2+ \dfrac rn \right) - \dfrac 1n \sum_{r=1}^n \ln \left(1 + \dfrac rn \right) \right] } \\ \displaystyle &= \exp{ \left[ \int_0^1 \ln(2+x) \,dx - \int_0^1 \ln(1+x) \,dx \right] } & \small\color{#20A900}{\text{By Riemann sums}}\\ \displaystyle &= \exp{ \left( \ln \dfrac{27}{16} \right) } \\ \displaystyle &= \boxed{\dfrac{27}{16}} \end{aligned}$