For all integers \(n>2\), prove \[ \frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^n}. \]
Solution
1. One has\[ \log(n!) = \sum_{k=1}^n\log k. \]
Since \(n!=\prod_{k=1}^n k\) and every factor is positive, the logarithm of the product is the sum of the logarithms.
2. For every integer \(n>1\),\[ \int_1^n\log x\,dx < \sum_{k=1}^n\log k. \]
For each \(k=2,\ldots,n\), the function \(\log x\) is strictly increasing on \([k-1,k]\), so \[ \int_{k-1}^k\log x\,dx < \log k. \] Summing these inequalities gives \[ \int_1^n\log x\,dx < \sum_{k=2}^n\log k = \sum_{k=1}^n\log k, \] because \(\log1=0\).
3. For every integer \(n\geq1\),\[ \sum_{k=1}^n\log k < \int_1^{n+1}\log x\,dx. \]
For each \(k=1,\ldots,n\), strict monotonicity gives \[ \log k < \int_k^{k+1}\log x\,dx. \] Indeed, \(\log x>\log k\) for every \(x\in(k,k+1]\). Summing over \(k\) yields the displayed inequality.
4. Therefore\[ n\log n-n+1 < \log(n!) < (n+1)\log(n+1)-n. \]
By steps <1>1–<1>3, \[ \int_1^n\log x\,dx < \log(n!) < \int_1^{n+1}\log x\,dx. \] Using \[ \int\log x\,dx=x\log x-x+C \] gives \[ \int_1^n\log x\,dx = n\log n-n+1 \] and \[ \int_1^{n+1}\log x\,dx = (n+1)\log(n+1)-n. \]
5. For every integer \(n>2\),\[ \boxed{ \frac{n^n}{e^{n-1}} < n! < \frac{(n+1)^{n+1}}{e^n} }. \]
The exponential function is strictly increasing. Exponentiating step 4 gives \[ e^{n\log n-n+1} < n! < e^{(n+1)\log(n+1)-n}. \] The two endpoint expressions simplify to \[ e^{n\log n-n+1} = \frac{n^n}{e^{n-1}} \] and \[ e^{(n+1)\log(n+1)-n} = \frac{(n+1)^{n+1}}{e^n}. \]
6. Q.E.D.
Step 5 is the required inequality.