# Why is $\sum \frac{1}{\ln(n)^{\ln n}}$ convergent?

Why is $$\sum \frac{1}{\ln(n)^{\ln n}}$$ convergent?

I tried Cauchy's condensation test along with other test but nothing works well on this series. Any ideas?

• Hint: $(\log(n))^{\log(n)} = n^{\log(\log(n))}$ . – Dan Brumleve Jan 20 '18 at 18:09

$\log(n)$ is unbounded, in particular for $N$ big enough $\log(n) > e^2$ for all $n\ge N$. Then
$$\sum_{n=N}^\infty \frac{1}{\log(n)^{\log(n)}} < \sum_{n=N}^\infty \frac{1}{e^{2\log(n)}} = \sum_{n=N}^\infty \frac{1}{n^2}$$
Rewrite the general term as $$\frac1{(\ln n)^{\ln n}}=\frac1{e^{\ln n\,\ln(\ln n)}}=\frac1{n^{\ln(\ln n)}}.$$ Now $\dfrac 1{n^{\scriptstyle \ln(\ln n)}}=o\Bigl(\dfrac1{n^r}\Bigr)$ for any $r>0$. In particular it is $\;o\Bigl(\dfrac1{n^2}\Bigr)$, which converges.