A new inequality has emerged! - Part 3

Let us use Induction to prove $\Sigma_{i=2}^{n} \frac{ln(i)}{i+1} < \frac{n^2}{4}, n \ge 2$ :

CASE I ( $n=2$ ): $\frac{ln (2)}{2+1} < \frac{2^2}{4} \Rightarrow \frac{ln (2)}{3} < 1$ (TRUE).

CASE ( $n = k$ ): $\Sigma_{i=2}^{k} \frac{ln (i)}{i+1} < \frac{k^2}{4}$ (ASSUMED TRUE).

CASE III ( $n = k+1$ ): $\frac{ln(k+1)}{(k+1)+1} + \Sigma_{i=2}^{k} \frac{ln (i)}{i+1} < \frac{k^2}{4} + \frac{ln(k+1)}{(k+1)+1}$ ;

or $\Sigma_{i=2}^{k+1} \frac{ln(i)}{i+1} < [\frac{k^2}{4} + \frac{ln(k+1)}{(k+1)+1}] \cdot \frac{(k+1)^2}{4} \cdot \frac{4}{(k+1)^2};$

or $\Sigma_{i=2}^{k+1} \frac{ln(i)}{i+1} < [\frac{k^2}{(k+1)^2} + \frac{4ln(k+1)}{(k+2)(k+1)^2}] \cdot \frac{(k+1)^2}{4};$

or $\Sigma_{i=2}^{k+1} \frac{ln(i)}{i+1} < [1 + \frac{4ln(k+1)}{k+2} - \frac{2 + 4ln(k+1)}{k+1} + \frac{1+4ln(k+1)}{(k+1)^2}] \cdot \frac{(k+1)^2}{4}$ ;

or $\Sigma_{i=2}^{k+1} \frac{ln(i)}{i+1} < f(k) \cdot \frac{(k+1)^2}{4}$ .

Since $\lim_{x \rightarrow \infty} f(x) \rightarrow 1$ AND $f'(x) > 0$ for all $x \ge 2$ (i.e. a strictly increasing function over this interval) , we conclude $\Sigma_{i=2}^{k+1} \frac{ln(i)}{i+1} < \frac{(k+1)^2}{4}$ (TRUE).

$\mathbb{Q.} \mathbb{E.} \mathbb{D.}$