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I need to prove that $a_n\ge a_{n+1}$, where $a_n=\frac{\ln\left(n\right)}{n}$ How can I do it?

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  • $\begingroup$ This is true for $n\geq 3$. $\endgroup$ – bjorn93 Jan 10 at 16:02
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If $$ f(x) = \frac{{\ln (x)}} {x} $$ then $$ f'(x) = \frac{{1 - \ln (x)}} {{x^2 }} $$ This means that $f$ is decreasing and this prove what you want. which is negative for $x \geq e$ .

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If you differentiate $\frac{\ln x}{x}$, you will see that it is decreasing for $x> e$.

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