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

Jan 20, 2017

The series:

${\sum}_{= 1}^{\infty} {\left(- 1\right)}^{n} {n}^{- \frac{1}{n}}$

is not convergent.

#### Explanation:

As the series:

${\sum}_{= 1}^{\infty} {\left(- 1\right)}^{n} {n}^{- \frac{1}{n}}$

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

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} {a}_{n}$ is convergent if:

${\lim}_{n \to \infty} {a}_{n} = 0$
${a}_{n + 1} < {a}_{n}$ at least for $n > {N}_{0}$

We have that:

${\lim}_{n \to \infty} {n}^{- \frac{1}{n}} = {\lim}_{n \to \infty} {\left({e}^{\ln} n\right)}^{- \frac{1}{n}} = {\lim}_{n \to \infty} {e}^{- \ln \frac{n}{n}} = 1$

so the series is not convergent.