Question

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

Answer 1

The series:

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

is not convergent.

Explanation:

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.