How do you test for convergence of $\infty \sum n = 2 {ln n}^{- ln n}$ $sum_(n=2)^(oo) lnn^(-lnn)$ ?

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Answer 1 · 2017-05-26T20:12:17

$\infty \sum n = 2 ln (n^{- ln n}) = - \infty$ $sum_(n=2)^oo ln(n^(-ln n)) = -oo$

Based on the properties of logarithms:

$ln (n^{- ln n}) = (- ln n) (ln n) = - {(ln n)}^{2}$ $ln(n^(-ln n)) = (-ln n) (ln n) = -(ln n)^2$

This means that:

$lim_(n->oo) a_n = lim_(n->oo) -(lnn)^2 = -oo$

The series does not satisfy Cauchy's necessary condition:

$lim_(n->oo) a_n = 0$

so it is not convergent.

As the terms are all negative it is divergent and:

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

How do you test for convergence of sum_(n=2)^(oo) lnn^(-lnn)∞∑n=2lnn−lnnsum_(n=2)^(oo) lnn^(-lnn)?