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#