Evaluate $\lim_{n\to\infty}\frac{12^n\cdot(n!)^5}{(n^2+n)^{2n}\cdot(n+1)^n}$

Question

Evaluate the following limit: $$\lim_{n\to\infty}\frac{12^n\cdot(n!)^5}{(n^2+n)^{2n}\cdot(n+1)^n}$$

The factorial function is creating a problem for me here. I can manage all other terms (by clubbing them together) but what to do with the factorial function.

I also thought of taking logarithm on both sides after assuming the limit as $y$ but that didn't help me much.

Any help is greatly appreciated.

Answer 1

The Stirling formula can be avoided. We will rely on elementary methods. We have $$b_n:=\frac{12^n\cdot(n!)^5}{(n^2+n)^{2n}\cdot(n+1)^n}\le {12^n(n!)^5\over n^{5n}}=\left [{12^{n/5}n!\over n^n}\right]^5=:a_n^5$$ Next $${a_{n+1}\over a_n}=12^{1/5}\left (1+{1\over n}\right )^{-n}\le 12^{1/5}2^{-1}<16^{1/5}2^{-1}=2^{-1/5}$$ Therefore $a_n\to 0,$ which implies $b_n\to 0.$ On the way we have used $$\left (1+{1\over n}\right )^n\ge 1+n{1\over n}=2$$ Remark The Stirling formula can be useful if we replace $12$ with $e^5.$ Indeed analyzing the proof we observe that ${a_{n+1}\over a_n}\to 1,$ which is inconclusive. If we apply the Stirling formula then $$a_n={e^nn!\over n^n}\approx \sqrt{2\pi n}\to \infty$$ and $${ a_n^5\over b_n}=\left (1+{1\over n}\right )^{3n}\to e^3$$ Hence $b_n\to \infty.$

Answer 2

Stirling's approximation is an asymptotic expression for the factorial, $$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac{1}{n}\right)\right).$$ We can safely ignore the polynomial terms; giving us asymptotically $$\frac{12^nn^{5n}e^{-5n}}{n^{5n}}\to0$$

Answer 3

As suggested in the comments, you can use Stirling's approximation: $$\frac{12^n(n!)^5}{(n^2+n)^{2n}(n+1)^n}\sim \frac{12^nn^{5n}e^{-5n}(2\pi n)^{5/2}}{(n+1)^{3n}n^{2n}}=\frac{12^n(2\pi n)^{5/2}}{e^{5n}}\left(\frac{n}{n+1}\right)^{3n}\to 0 $$