To test the convergence $\sum{\frac{1}{(\ln {n})^{\ln{n}}}}$

Question

How to test the convergence of following series $$\sum{\frac{1}{(\ln {n})^{\ln{n}}}}$$

I have tried Cauchy condensation test and gives me nothing

Answer 1

Method 1: Claim: $(\ln n)^{\ln n} > n^2$ for $n>e^{e^2}.$ The claim implies

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

for these $n.$ Since $\sum 1/n^2$ converges, so does $\sum 1/(\ln n)^{\ln n}.$

To prove the claim, apply $\ln $ to both sides. I'll leave this to the reader for now.

Method 2: Cauchy condensation: Consider the series

$$\tag 1\sum \frac{2^n}{(\ln 2^n)^{\ln 2^n}} = \sum \frac{2^n}{(n\ln 2)^{n\ln 2} }= \sum \frac{2^n}{((n\ln 2)^{\ln 2})^n}.$$

Now $(n\ln 2)^{\ln 2}\to \infty,$ so eventually $(n\ln 2)^{\ln 2} \ge 4.$ This shows the terms in $(1)$ are eventually bounded above by $2^n/4^n = 1/2^n.$ This proves convergence of the original series.

Answer 2

Let's try that Cauchy condensation test again with a ratio test:

$$\frac1{(\ln n)^{\ln n}}\implies\frac{2^n}{(\ln2^n)^{\ln2^n}}=\frac{2^n}{(cn)^{cn}}\implies\frac{(2/c(n+1))^{c(n+1)}}{(2/cn)^{cn}}\to0$$

where $c=\ln2$.

Thus, it converges.