Limiting factorials

Let $\displaystyle P =\lim_{ n \rightarrow \infty }{ \left( \dfrac{ 5n \choose 3n }{ 3n \choose 2n } \right) }^{ \small \dfrac{1}{n} }$ We notice that $\displaystyle \dfrac{ 5n \choose 3n }{ 3n \choose 2n } = \prod_{ r=0 }^{ 2n-1 }\dfrac{ 5n-r }{ 3n-r }$ .So $\displaystyle P = \lim_{ n \rightarrow \infty }{ \left( \prod_{ r=0 }^{ 2n-1 }\dfrac{ 5n-r }{ 3n-r } \right) }^{ \dfrac{1}{n} }$ . Now taking log on both sides. $\displaystyle \log {P} = \lim_{ n \rightarrow \infty } \dfrac{ 1 }{ n } \log \prod_{ r=0 }^{ 2n-1 }\dfrac{ 5n-r }{ 3n-r }$ Using $\log ab = \log a +\log b$ . We get $\displaystyle \log P = \lim_{ n \rightarrow \infty } \dfrac{ 1 }{ n }\sum_{ r=0 }^{ 2n-1 }\log{ \left( \dfrac{ 5n-r }{ 3n-r } \right) } \rightarrow \log P = \int_0^2 \log{ \left( \dfrac{ 5-x }{ 3-x } \right) }dx$

Hence we get $\displaystyle P = \dfrac{ { 5 }^{ 5 } }{ { 3 }^{ 6 } }$

So $a=5$ & $b=3$ . Hence $\boxed {\boxed{ a+b=8}}$

alternatively write $L=\lim_{n\to \infty} \left( \dfrac{\Gamma(5n+1)\Gamma(n+1)}{\Gamma^2(3n+1)} \right)^{1/n}$ we use: $\ln(L)=\lim_{n\to \infty}\left(\dfrac{\ln(\Gamma(5n+1))-2\ln(\Gamma(3n+1))+\ln(\Gamma(n+1))}{n}\right)=^{l'opital} \lim_{n\to \infty} 5\psi(5n+1)-6\psi(3n+1)+\psi(n+1)$ we use tha fact as asymptopic levels $\psi(n+1)=-\gamma+H_n\approx \ln(n)$ $\ln(L)= \lim_{n\to \infty} 5\ln(5n)-6\ln(3n)+\ln(n)\to L=\dfrac{5^5}{3^6}$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left[ \frac{5n \choose 3n }{3n \choose 2n } \right ] ^{1/n}$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left[ \frac{(5n)!}{(3n)! (2n)!} \cdot \frac{(2n)! n!}{(3n)!}\right ] ^{1/n}$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left[ \frac{(5n)! n!}{(3n)! (3n)!}\right ] ^{1/n}$

By Stirling's approximation :

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left[ \frac{\sqrt{10 \pi n } (5n/e)^{5n} \cdot \sqrt{2 \pi n } (n/e)^{n}}{\sqrt{6 \pi n } (3n/e)^{3n} \cdot \sqrt{6 \pi n } (3n/e)^{3n}} \right ] ^{1/n}$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left[ \sqrt{\frac{20}{36}} \cdot \frac{5^{5n}}{3^{6n}} \right] ^{1/n}$

$\large \displaystyle L = \lim_{n \rightarrow \infty} \left[ \sqrt{\frac{20}{36}} \right]^{1/n} \cdot \lim_{n \rightarrow \infty} \left [ \frac{5^{5n}}{3^{6n}} \right] ^{1/n}$

$\color{#20A900} \boxed{ \large \displaystyle L = \frac{5^5}{3^{2 \cdot 3}} }$

Thus:

$\color{#3D99F6} \large \displaystyle a=5, b=3, \boxed{\large \displaystyle a+b=8 }$