How to prove that the sequence is decreasing $a_{n}=\frac{ln(n)}{n^2}$

Question

Is my way/proof good and completely mathematically rigorous?

$a_{n}=\frac{ln(n)}{n^2}$ --> $a_{n+1}=\frac{ln(n+1)}{(n+1)^2}$

$\frac{ln(n)}{n^2} > \frac{ln(n+1)}{(n+1)^2}$

$(n+1)^2ln(n)>n^2ln(n+1)$

$ne^{(n+1)^2} > (n+1)e^{n^2}$

$\frac{n}{n+1} > e^{n^2-(n+1)^2}$

$\frac{n}{n+1} > e^{-(2n+1)}$

$\frac{n}{n+1} > \frac{1}{e^{2n+1}}$

The left hand side is increasing and the right hand side is decreasing and there for the inequality holds for every $n > 1$ and the initial sequence is decreasing

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I got an inspiration from @Alex

$a_{t}=\frac{t}{e^{2t}}$ --> $a_{t+1}=\frac{t+1}{e^{2t+2}}$

$a_{t} > a_{t+1}$

$\frac{t}{e^{2t}} > \frac{t+1}{e^{2t+2}}$

$e^{2t}e^2t > e^{2t}(t+1)$

$e^2t > t+1$

$e^2t-t > 1$

$t > \frac{1}{e^2-1}$

Answer 1

You're close, but it's easier. Set $\log n = t$, then $a_t = \frac{t}{e^{t^2}}$. Consider $$ \frac{a_{t+1}}{a_t} = \frac{t+1}{e^{(t+1)^2}} \cdot \frac{e^{t^2}}{t} = \bigg(1+\frac{1}{t}\bigg)e^{-2t -1} \leq 2 e^{-2t -1}<1 \ \forall \ t>\frac{\log 2 -1}{2}>0 $$ You can check the last inequality by noting that $e^{-2t-1}$ is decreasing in $t$.

Answer 2

Hint

Consider $$f(x)=\frac{\log(x)}{x^2}$$ So $$f'(x)=\frac{1-2 \log (x)}{x^3}$$ which is negative for $x \geq 2$.