How do you test the series $Sigma (n^n)/(lnn)^n$ from $n=[2,oo)$ by the ratio test?

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Answer 1 · 2017-06-29T22:09:54

$sum_(n=2)^oo n^n/(lnn)^n = oo$

It's really easier in this case to use the root test:

$a_n = n^n/(lnn)^n = (n/lnn)^n$

So that:

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

Thus the series is not convergent, and as it has positive terms, it is divergent:

$sum_(n=2)^oo n^n/(lnn)^n = oo$

How do you test the series Sigma (n^n)/(lnn)^nSigma (n^n)/(lnn)^n from n=[2,oo)n=[2,oo) by the ratio test?