How do you test for convergence given #Sigma (-1)^n n^(-1/n)# from #n=[1,oo)#?

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Answer 1 · 2017-01-20T11:49:30

The series:

#sum_(=1)^oo (-1)^n n^(-1/n)#

is not convergent.

As the series:

#sum_(=1)^oo (-1)^n n^(-1/n)#

is alternating, we can test it for convergence using Leibniz's criteria stating that:

#sum_(n=1)^oo (-1)^n a_n# is convergent if:

# lim_(n->oo) a_n = 0#
# a_(n+1) < a_n# at least for #n> N_0#

We have that:

# lim_(n->oo) n^(-1/n) = lim_(n->oo) (e^lnn)^(-1/n)= lim_(n->oo) e^(-lnn/n) = 1#

so the series is not convergent.