To check whether given series is convergent or divergent. Given series is 
$$\sum_{n=2}^{\infty}\left(\frac{1}{\ln(n)}\right)^{\ln(n)}$$.
I have attempted as follows but not sure.
Making use of this inequality $\ln(n)^{\ln(n)}$$<$n..and then by using comparison test and p-test I think series should be divergent..But not sure..
 A: With @Qiaochu Yuan rewriting,   note that, for instance,
$$\frac1{\log n^{\log n}}=\frac1{n^{\log(\log n)}}<\frac1{n^2}$$
if $n$ is large enough, and the latter converges.
A: The general term is decreasing (at least, for $n\ge 3$). Then you can use Cauchy condensation test (with $3^k$ in place of $2^k$, it works the same). Your series has the same behaviour of
$$\sum_k 3^k\left(\frac{1}{k\ln(3)} \right)^{k\ln(3)}\,.$$
The general term can be bounded from above by
$$3^k\left(\frac{1}{k\ln(3)} \right)^{k\ln(3)} \le \left(\frac{3}{k} \right)^{k}$$
A: It is the other way around:
$$\left(\ln n\right)^{\ln n}>n \iff \ln n \cdot \ln {\ln n}>\ln n \iff \frac{1}{\left(\ln n\right)^{\ln n}}<\frac1n,$$
However, this is not sufficient, because: $\frac{1}{n\ln n}<\frac1n$, but both diverge.
Use instead:
$$\left(\ln n\right)^{\ln n}>n^2 \iff \ln n \cdot \ln {\ln n}>2\ln n \iff \ln \ln n>2\iff $$
$$\ln n >e^2 \iff n> e^{e^2} \iff \frac{1}{\left(\ln n\right)^{\ln n}}<\frac{1}{n^2} \Rightarrow \text{converge}.$$
