Factorials at infinity

We Can Also use Stirling's Approximation, i.e. $n! \approx \sqrt{2\pi n} \left( \dfrac{n}{e} \right)^{n}$ to make it easier

Let l(m)= $\lim _{ n\rightarrow \infty }{ { (\frac { (mn)! }{ { n! }^{ m } } })^{ \frac { 1 }{ n } } }$

Divide numerator and denominator by n^(mn) and take log on both sides to obtain,

log l(m)= $\lim _{ n\rightarrow \infty }{ \frac { 1 }{ n } \left( \sum _{ r=1 }^{ mn }{ \log { \frac { r }{ n } } } -m\sum _{ r=1 }^{ n }{ \log { \frac { r }{ n } } } \right) }$

We can write this as log l(m)= $\int _{ 0 }^{ m }{ \log { (x)dx\quad -\quad m } } \int _{ 0 }^{ 1 }{ \log { (x)dx } }$

After evaluating this integral,

log l=m log m

so l= ${ m }^{ m }$

required ans=l(2)+l(3) =2^2+3^3=31