Exercise 8.1.2

Obtain asymptotically tight bounds on $\lg(n!)$ without using Stirling's approximation. Instead, evaluate the summation $\sum_{k=1}^n\lg{k}$ using techniques from section A.2.

First we show that it is $\O(n\lg{n})$:

$$\sum_{k=1}^n\lg{k} \le \sum_{k=1}^n\lg{n} = n\lg{n} = \O(n\lg{n})$$

Next we show that it is $\Omega(n\lg{n})$:

\begin{align} \sum_{k=1}^n\lg{k} &= \sum_{k=1}^{\lfloor n/2 \rfloor}\lg{k} + \sum_{k=\lfloor n/2 \rfloor + 1}^n\lg{k} \\ &\ge \sum_{k=\lfloor n/2 \rfloor + 1}^n\lg{k} \\ &\ge \sum_{k=n/2}^n\lg{k} \\ &\ge \sum_{k=n/2}^n\lg{n/2} \\ &\ge (n/2)\lg(n/2) \\ &= \frac{1}{2}n\lg{n} - \frac{1}{2}n \\ &= \Omega(n\lg{n}) \end{align}