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How do you test for convergence of #Sigma (ln(n))^-n# from #n=[2,oo)#?

1 Answer

Mar 9, 2017

The series:

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

is convergent.

Explanation:

The series has positive terms so we can use the root test and we have:

#L= lim_(n->oo) root(n)(a_n) = lim_(n->oo) root(n)((ln(n)^(-n))) = lim_(n->oo) 1/lnn = 0#

As #L < 1# the series is convergent.

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