Calculation of $$\lim_{n\rightarrow\infty}\frac{3^{3n}\cdot (n!)^3}{(3n+1)!}=$$
$\bf{My\; Try::}$ Using Stirling Approximation $\displaystyle (n!\approx\left(\frac{n}{e}\right)^n\sqrt{2\pi n})$,We get
Limit $$l=\lim_{n\rightarrow\infty}\frac{3^{3n}\cdot \left(\frac{n}{e}\right)^{3n}\left(\sqrt{2\pi n}\right)^3}{\left(\frac{3n+1}{e}\right)^{3n+1}\sqrt{2\pi (3n+1)}} = \frac{2\pi}{3\sqrt{3}}$$
My question is how can we solve it Using Reinman sum (Limit as a sum) or any other method
Help me ,Thanks
First of all, let's see what the result should be for general $k$.
$\begin{array}\\ f(k, n) &=\dfrac{k^{kn} (n!)^k}{(kn)!}\\ &\approx \dfrac{k^{kn} (n^n\sqrt{2\pi n}e^{-n})^k}{(kn)^{kn}\sqrt{2\pi kn}e^{-kn}(kn)!}\\ &= \dfrac{k^{kn} n^{kn}(2\pi)^{k/2} n^{k/2}e^{-kn}}{k^{kn}n^{kn}\sqrt{2\pi kn}e^{-kn}}\\ &= \dfrac{(2\pi)^{(k-1)/2} n^{(k-1)/2}}{\sqrt{ k}}\\ \end{array} $
As a check, this gives $f(3, n) \approx \dfrac{2\pi n}{\sqrt{ 3}} $ which agrees with your result, since $(k-1)/2 = 1$ for $k=3$.
Note that the $(3n+1)!$ was sort of a fake, since $3n(3n)!$ would have given the same limit.
If you want to so some sort of Riemann sum, you would have to take logs and use $\ln(n!) =\sum_{i=1}^n \ln(i) \approx \int_{i=1}^n \ln(x)dx =(x \ln(x)-x)\big|_0^n =n\ln(n)-n $. This is quite close to Stirling's $\ln(n!) \approx n\ln(n)-n+\frac12(\ln(n)+\ln(2\pi)) $. The reason is that if $f$ is monotonic then the error in using $\sum_{i=1}^n f(i) \approx \int_{i=1}^n f(x)dx $ is bounded by $f(0)+f(n)$ which is, in this case, $\ln(n)$. This can be proved by using $\min(f(n), f(n+1)) \le \int_n^{n+1} f(x) dx \le \max(f(n), f(n+1)) $.
What you get for $\ln f(k, n)=g(k, n)$ is
$\begin{array} fg(k, n) &\approx kn\ln(k)+k\ln(n!)-\ln((kn)!)\\ &=kn\ln(k)+k(n\ln(n)-n)-(kn\ln(kn)-kn)\\ &=kn\ln(k)+kn\ln(n)-kn-(kn(\ln(k)+\ln(n))-kn)\\ &=kn\ln(k)+kn\ln(n)-kn-(kn(\ln(k)+\ln(n))-kn)\\ &=0\\ \end{array} $
The reason for this is that the error in $\ln(n!) \approx n \ln(n)-n $ is of order $\ln(n)$, so the error in $\ln(n!^k)$ is of order $k\ln(n) =\ln(n^k) $ which fits the precise result nicely.
In other words, you would have to use a more precise approximation to $\sum_{i=1}^n \ln(i)$ and the best you could do would be the same as Stirling's approximation with the constant ($\sqrt{2\pi}$) being undetermined.
More precisely, if you used $\ln(n!) \approx n\ln(n)-n+\frac12 \ln(n)+c $, you would get $\ln(f(k, n)) \approx (k-1)c+\frac{k-1}{2}\ln(n)-\frac12 \ln(k) $.
And that's about all I can think of.
