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

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Answer 1 · 2017-05-14T14:33:15

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

$sum_(n=1)^oo ln^(-ln(n))(n) = "undefined"$

$sum_(n=2)^oo ln^(-ln(n))(n) ~~ 5.7$

Answer 2 · 2017-05-14T17:12:45

The series is convergent.

For $n > 1$ there exists $lambda_n$ such that

$log_e n = e^(lambda_n)$

and also

$(log_e n)^(-log_e n) = (e^(lambda_n))^(-log_e n)=e^(-lambda_n log_e n) = 1/n^(lambda_n) = 1/n^(log_e(log_en))$

We know also that for $n > e^(e^2) approx 1618->log_e(log_e n)> 2$ so this series is convergent because the series

$sum_(k=1)^oo 1/n^2$ is convergent.

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