Htydjtk

I have this problem nowadays:

$$lim_{xto{infty}}frac{k^{x}x!}{x^x}$$

I found that it diverges when $k$ is greater than $e$ and converges when smaller than $e$.

I want to know the reason!! Please help me!!

It's simple.

Apply Stirling approximation to the factorial:

$$x! approx sqrt{2pi x} x^x e^{-x}$$

And substitute it into your limit. You simplify a bit and arranging you get

$$lim_{xto +infty} frac{k^x}{e^x}sqrt{2pi x}$$

Actually the $sqrt{2pi x}$ is quite irrelevant. What you notice is the previous term, which you can also write as

$$lim_{xto +infty} left(frac{k}{e}right)^x$$

It's rather obvious now, that if $k/e < 1$ the limit converges, and if $k/e >1$ it does diverge. Whence your claim.

Nota: The square root term is irrelevant for if $k/e > 1$ it does diverge no matter the root. if $k/e < 1$ it does converge also in virtue of the l'Hôpital rule, and the root becomes irrelevant.

Nota2: Stirling Approximation works well if indeed $x$ goes to the infinity. Actually it works quite well even for small values of $x$. Hence your safe.

Hint (without Stirling). Consider the sequence $a_n=frac{n!k^n}{n^n}$ and find the limit
$$lim_{nto+infty}frac{a_{n+1}}{a_n}=lim_{nto+infty}frac{(n+1)!k^{n+1}}{(n+1)^{n+1}}cdot frac{n^n}{n!k^n}=kcdot lim_{nto+infty}frac{1}{(1+1/n)^n}$$
then use the ratio test for sequences (see Proof attempt to the ratio test for sequences).

Very similar to Von Neumann's answer.

$$y=frac{k^{x}x!}{x^x}implies log(y)=x log(k)+log(x!)-xlog(x)$$ Using Stirling, we have
$$log(y)=x left(log (k)-1right)+frac{1}{2}
left(log (2 pi )+log left({x}right)right)+frac{1}{12
x}+Oleft(frac{1}{x^3}right)$$ $$log(y)sim x left(log (k)-1right)+frac{1}{2}
left(log (2 pi )+log left({x}right)right)$$ which makes
$$y sim sqrt{2 pi x }, e^{x (log (k)-1)}=sqrt{2 pi x },e^{xlog(frac k e)}$$