# How do you determine the convergence or divergence of Sigma ((-1)^(n+1)n)/(2n-1) from [1,oo)?

Dec 23, 2016

The series:
${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} \frac{n}{2 n - 1}$
is divergent.

#### Explanation:

You can determine whether an alternating series converges using Leibniz' criteria, which states that:

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

converges if:

(i) ${a}_{n} > {a}_{n + 1}$
(ii) ${\lim}_{n} {a}_{n} = 0$

As the general term of the series above can be expressed as:

${a}_{n} = - \frac{n}{2 n - 1}$

We can quickly see that:

${\lim}_{n} {a}_{n} = {\lim}_{n} - \frac{n}{2 n - 1} = {\lim}_{n} - \frac{1}{2 - \frac{1}{n}} = - \frac{1}{2}$

so that condition (ii) is not met and the series is divergent.