# How do you test the alternating series Sigma (-1)^n/rootn n from n is [2,oo) for convergence?

Feb 13, 2017

The alternating series:

${\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} / \sqrt[n]{n}$

is not convergent.

#### Explanation:

This is an alternating series so we can apply Leibniz's test stating that the series is convergent if given:

a_n = 1/(root(n)(n)

we have:

(i) ${\lim}_{n \to \infty} {a}_{n} = 0$

(ii) ${a}_{n + 1} / {a}_{n} < 1$

Now consider the sequence:

${b}_{n} = \ln {a}_{n} = \ln \left(\frac{1}{\sqrt[n]{n}}\right) = - \frac{1}{n} \ln n$

We have:

${\lim}_{n \to \infty} {b}_{n} = 0$

And:

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} {e}^{{b}_{n}}$

but as ${e}^{x}$ is a continuous function:

${\lim}_{n \to \infty} {a}_{n} = {e}^{{\lim}_{n \to \infty} {b}_{n}} = {e}^{0} = 1$

The Leibniz's test is then not satisfied and the series is not convergent.