Chapter 4, Problem 53C

Explanation of Solution

Given:

It is given that ${\displaystyle \sum _{i=1}^n\log i}=\Omega \left(n\log n\right)$.

Proof:

Let us prove the above statement by the following assumption that “n” is even number of integers:

$\begin{array}{c}{\displaystyle \sum _{i=1}^n{\log }_{2}i}\ge {\displaystyle \sum _{i=n/2}^n{\log }_2}i\\ {\displaystyle \sum _{i=n/2}^n{\log }_2}i={\displaystyle \sum _{i=n/2}^n{\log }_{2}n/2}\\ ={\displaystyle \sum _{i=n/2}^n\left({\log }_{2}n\right)-\left({\log }_{2}2\right)}(Note:log2=1)\end{array}$

$\begin{array}{c}={\displaystyle \sum _{i=n/2}^n({\log }_2}\end{array}$