Computing the $\lim_{n\to\infty} n^{\frac{1}{n!}}$?

Question

I want to find the following limit

$$\lim\limits_{n\rightarrow \infty} n^{\frac{1}{n!}}$$

I tried solution as follows:

Let $L=n^{\frac{1}{n!}}$ this implies $\log L=\frac{\log n}{n!}$ which is $\frac{\infty}{\infty}$ form as $n\rightarrow \infty.$ Don't know how to proceed. Help required

Answer 1

Hint:

$$1 \le n^{1/n!} \le (n!)^{1/n!}$$

Answer 2

HINT:

$$\lim_{n\to\infty}n^{\frac{1}{n!}}=\lim_{n\to\infty}\exp\left(\ln\left(n^{\frac{1}{n!}}\right)\right)=\lim_{n\to\infty}\exp\left(\frac{1}{n!}\ln\left(n\right)\right)=\exp\left(\lim_{n\to\infty}\frac{\ln(n)}{n!}\right)$$

Answer 3

I am writing solution on the hint provided by @Crostul

We have $0<\frac{\log n}{n!}<\frac{1}{(n-1)!}.$ Therefore $0\leq \lim\limits_{\lim\rightarrow\infty} \frac{\log n}{n!}\leq 0$ that is $L=1.$

Answer 4

Try with

$$n^{\frac{1}{n!}} = e^{\frac{1}{n!}\ln(n)}$$

Then your limit will be in the form $\frac{\infty}{\infty}$. The $\ln(n)$ function grow reeeeeeeeeeally slow with respect to $n!$ thus the fraction is zero and you get $e^0 = 1$.

Limiti is $1$.