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

Question

4050 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-01T05:42:54

The series converges (absolutely)

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)

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