Question

Show that logn! Is greater than $\mathbf{}\frac{\mathbf{n}\mathbf{}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{n}}{\mathbf{4}}$for n > 4 . [Hint: Begin with the inequality$\mathit{n}\mathbf{!}\mathbf{>}\mathit{n}\mathbf{(}\mathit{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathit{n}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{\dots }\mathbf{.}\mathbf{[}\mathit{n}\mathbf{/}\mathbf{2}\mathbf{]}\mathbf{.}\mathbf{]}$

Short answer

Using the definition of factorials, it is proved that $\log n!>\frac{n\log n}{4}$ for n > k and k = 4

Step-by-step solution

Expanding the factorial:

$$\begin{gathered} ifk=4withn>k \\ \begin{array}{r}\log n!=\log n!\\ \Rightarrow \log (n\cdot (n-1)\dots .3\cdot 2\cdot 1)\\ \Rightarrow \log n+\log (n-1)+\dots +\log 3+\log 2+\log 1\\ \Rightarrow \log n!\ge \log n+\log (n-1)+\dots +\log \frac{n}{2}\end{array} \end{gathered}$$

Simplifying the inequality for n > 4 :

Since $\frac{n}{2}\le i$whenever $i\in \left\{\frac{n}{2},\frac{n}{2}+1,\dots ,n-1\right\}$

role="math" localid="1668571592739" $\begin{array}{r}\log n!\ge \log \frac{n}{2}+\log \frac{n}{2}+\dots +\log \frac{n}{2}\\ \Rightarrow \log n!\ge \frac{n}{2}\log \frac{n}{2}\\ \log \frac{n}{2}>\frac{1}{2}\log n\text{for}n>4\\ \log n!\ge \frac{n}{4}\log n\end{array}$

Therefore, it is proved that $\log n!>\frac{n\log n}{4}$ for n > k and k = 4 .