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

Question

3489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-10T01:34:08

The series:

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

is not convergent.

A necessary condition for the series to converge is that:

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

For this series:

#a_n = (-1)^n n^(-1/n) = (-1)^n e^ln(n^(-1/n)) = (-1)^n e^(-lnn/n)#

so:

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

and as #e^x# is continuous:

#lim_(n->oo) abs (a_n) = e^(lim_(n->oo) (-lnn/n)) = e^0 = 1#

Consequently:

#lim_(n->oo) a_n !=0#

which means the series is not convergent.