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

Mar 9, 2017

The series:

${\sum}_{n = 2}^{\infty} \frac{n {\left(- 1\right)}^{n + 1}}{\ln} n$

is not convergent.

#### Explanation:

A necessary condition for any series to converge is that:

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

which also implies:

${\lim}_{n \to \infty} \left\mid {a}_{n} \right\mid = 0$

In our case:

${\lim}_{n \to \infty} \left\mid {a}_{n} \right\mid = {\lim}_{n \to \infty} \frac{n}{\ln} n = \infty$

so the series is not convergent.

We can also see that taking only the terms of even order:

${\lim}_{n \to \infty} {a}_{2 n} = {\lim}_{n \to \infty} \frac{2 n {\left(- 1\right)}^{2 n + 1}}{\ln} \left(2 n\right) = {\lim}_{n \to \infty} - \frac{2 n}{\ln} \left(2 n\right) = - \infty$

while for terms of odd order:

${\lim}_{n \to \infty} {a}_{2 n + 1} = {\lim}_{n \to \infty} \frac{\left(2 n + 1\right) {\left(- 1\right)}^{2 n + 2}}{\ln} \left(2 n + 1\right) = {\lim}_{n \to \infty} \frac{2 n + 1}{\ln} \left(2 n + 1\right) = + \infty$

so the series is irregular.