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.

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