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

Sep 1, 2017

The series converges (absolutely)

#### Explanation:

The series has positive ad negative elements, we check for conditional / absolute convergence.

Absolute convergence : $\sum {a}_{n}$ is absolutely convergent if $\sum | {a}_{n} |$ is convergent

Conditional convergence : $\sum {a}_{n}$ is conditionally convergent if $\sum | {a}_{n} |$ is divergent and $\sum {a}_{n}$ is convergent

|a_(n+1)|/|a_n|=|1/((n+1)!)|/(1/|(n!)|)=1/|(n+1)|

${\lim}_{n \to + \infty} \frac{1}{n + 1} = 0$

As limit $< 1$, the series converges (absolutely)