Convexity of $\frac{1}{n\log(n)}$ for $n>2$

Question

I am trying to show this sequence is convex, i.e. that $$a_n \leq \frac{a_{n+1}+a_{n-1}}{2}.$$

I have tried to show the sequence $$a_n - \left(\frac{a_{n+1}+a_{n-1}}{2}\right)\leq 0$$ directly.

Using the derivative to show the sequence is decreasing and that for $n=3$ it is already smaller than $0$.

Answer 1

HINT

By second derivative test for $$f(x)=\frac1{x\log x}$$

we obtain

$$f''(x)=\frac{2\log^2x+3\log x+2}{x^3\log^3 x}$$

Answer 2

$\frac{1}{x}$ is convex and $\log(x)$ is concave on $\mathbb{R}^+$, hence $\frac{1}{x}$ is log-convex on $\mathbb{R}^+$.
$\log\log x$ is concave on $(1,+\infty)$, hence $\frac{1}{\log x}$ is log-convex on $(1,+\infty)$.
It follows that $\frac{1}{x}\cdot\frac{1}{\log x}$ is log-convex on $(1,+\infty)$, hence convex.