Compute \[ \lim_{n\to\infty}\left(\frac{n^n}{n!}\right)^{1/n}. \]
Solution
For \(n\ge1\), set \[ a_n\coloneqq\left(\frac{n^n}{n!}\right)^{1/n}. \]
1. For every \(n\ge2\),\[ \int_1^n\log x\,dx \le \log(n!) \le \int_1^n\log x\,dx+\log n. \]
Since \(\log x\) is increasing, for \(2\le k\le n\), \[ \int_{k-1}^k\log x\,dx\le\log k. \] Summing gives \[ \int_1^n\log x\,dx\le\sum_{k=2}^n\log k=\log(n!). \]
Similarly, for \(1\le k\le n-1\), \[ \log k\le\int_k^{k+1}\log x\,dx. \] Hence \[ \log((n-1)!)\le\int_1^n\log x\,dx, \] and adding \(\log n\) gives the upper bound.
2. One has\[ 1-\frac{1+\log n}{n} \le \log a_n \le 1-\frac1n. \]
The integral in step 1 is \[ \int_1^n\log x\,dx =n\log n-n+1. \] Also, \[ \log a_n =\frac{n\log n-\log(n!)}{n}. \] Substituting the upper and lower bounds for \(\log(n!)\) from step 1 yields exactly the stated inequalities.
3. The requested limit is\[ \boxed{e}. \]
Both bounds in step 2 tend to \(1\), so the squeeze theorem gives \[ \lim_{n\to\infty}\log a_n=1. \] Continuity of the exponential function therefore gives \[ \lim_{n\to\infty}a_n=e. \]
4. Q.E.D.
Step 3 computes the required limit.