Question

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

Answer 1

The series converges (absolutely)

Explanation:

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

Absolute convergence : `suma_n` is absolutely convergent if `sum|a_n|` is convergent

Conditional convergence : `suma_n` is conditionally convergent if `sum|a_n|` is divergent and `suma_n` is convergent

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

`lim_(n->+oo)1/(n+1)=0`

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