# Does sum (ln n)^(-ln n) converge?

May 14, 2017

The sum converges, if you avoid the undefined term at $n = 1$

#### Explanation:

${\sum}_{n = 1}^{\infty} {\ln}^{- \ln \left(n\right)} \left(n\right) = \text{undefined}$

${\sum}_{n = 2}^{\infty} {\ln}^{- \ln \left(n\right)} \left(n\right) \approx 5.7$

May 14, 2017

The series is convergent.

#### Explanation:

For $n > 1$ there exists ${\lambda}_{n}$ such that

${\log}_{e} n = {e}^{{\lambda}_{n}}$

and also

${\left({\log}_{e} n\right)}^{- {\log}_{e} n} = {\left({e}^{{\lambda}_{n}}\right)}^{- {\log}_{e} n} = {e}^{- {\lambda}_{n} {\log}_{e} n} = \frac{1}{n} ^ \left({\lambda}_{n}\right) = \frac{1}{n} ^ \left({\log}_{e} \left({\log}_{e} n\right)\right)$

We know also that for $n > {e}^{{e}^{2}} \approx 1618 \to {\log}_{e} \left({\log}_{e} n\right) > 2$ so this series is convergent because the series

${\sum}_{k = 1}^{\infty} \frac{1}{n} ^ 2$ is convergent.