# How do you determine if the series the converges conditionally, absolutely or diverges given Sigma (-1)^(n+1)/(n+1) from [1,oo)?

Sep 19, 2017

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n + 1} / \left(n + 1\right) \text{ }$ is conditionally convergent.

#### Explanation:

First note that:

${\sum}_{n = 1}^{\infty} \frac{1}{n + 1} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} + \ldots$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} \frac{1}{n + 1}} > \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \ldots$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} \frac{1}{n + 1}} = \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \left(\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}\right) + \ldots$

$\textcolor{w h i t e}{{\sum}_{n = 1}^{\infty} \frac{1}{n + 1}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \ldots \text{ }$ diverges.

So our series is not absolutely convergent.

However, note that $\frac{1}{n + 1}$ is monotonically decreasing with limit $0$ as $n \to \infty$.

Hence:

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

is conditionally convergent.