Show without derivative that function $\frac{\ln{n}}{ n\ln{\ln{n}}}$ is decreasing

Question

I have a problem with showing the function $\displaystyle \frac{\ln{n}}{n \ln{\ln{n}}}$ is decreasing.

I came to form $(n+1)^{\ln{\ln{n}}}<(n)^{\ln{\ln{(n+1)}}}$ and I don't know how to show that this inequality holds from some $n_0$.

Answer 1

Assuming that $\log\log n$ is positive, $$a_{n+1}\triangleq\frac{\log(n+1)}{(n+1)\log\log(n+1)}< \frac{\log n+\frac{1}{n}}{(n+1)\log\log n},$$ so, in order to prove $a_{n+1}<a_n$, it is sufficient to show that: $$\frac{\log n+\frac{1}{n}}{(n+1)\log\log n}-\frac{\log n}{n\log\log n}\leq 0,$$ or: $$\log n+\frac{1}{n}-\left(1+\frac{1}{n}\right)\log n \leq 0,$$ or: $$1-\log n\leq 0$$ that is trivial given that $n>e$.

Answer 2

For $n>e$ $$\frac{\ln n}{n\ln \ln n}>\frac{\ln n}{n\ln\ln (n+1)}>\frac{\ln(n+1)}{(n+1)\ln\ln (n+1)}$$ Using the fact that $\frac{\ln n}{n}$ is decreasing for $n>e$ and hence $\frac{\ln n}{n}>\frac{\ln (n+1)}{(n+1)}$. Therefore $a_n >a_{n+1}$