Finding a positive lower bound of the sequence $\frac{\sqrt[n]{n!}}n$

Question

I am given a sequence $$\frac{\sqrt[n]{n!}}n.$$ Can I show that the sequence is bounded below by a real no. which is greater than 0, by not calculating the limit of it....???thank you

Answer 1

If I properly read the question, you are concerned by $$a_n=\frac{(n!)^{\frac{1}{n}}}{n}$$ So $$\log(a_n)=\frac 1n \log(n!)-\log(n)$$ As usual when working with factorials, Stirling approximation is useful. You will in particular find this very nice inequality $$\sqrt{2\pi}n^{n+\frac 12}e^{-n} \leq n! \leq en^{n+\frac 12}e^{-n}$$ which would give you good approximations of upper and lower bounds for $a_n$.

Answer 2

By comparing $\log n!$ with $\int_1^n \log x \, dx$, we get $$ n! \ge e\left(\frac ne\right)^n $$ and so $$ \frac{\sqrt[n]{n!}}n \ge \frac{\sqrt[n]{e}}e \ge \frac1e $$