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

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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.

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