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.