Prove $\sum_{j=1}^{n^2} \log_n(2j-1)\leq2n^2 $

Question

Deduce whether the statement is true or false.

Suppose $n\in \mathbb N \setminus \{0,1\}$. Then, $$\sum_{j=1}^{n^2} \log_n(2j-1)\leq2n^2 $$

I would like to ask what inequality I can apply or any hints on how to start the proof? Are Weierstrass’ Inequality, Cauchy-Schwarz Inequality or AM-GM Inequality useful? (I only learnt these inequalities recently) Thanks.

Answer 1

Another way to look at the question: You can use $\textbf{Jensen's Inequality}$ repetitively for each highest-lowest pair (since $\log_n$ is concave) to show in a relatively simple way:

$$\sum_{j=1}^{n^2} \log_n(2j-1)\leq \sum_{j=1}^{n^2} \log_n {n^2} = 2n^2$$