How do you test the series #Sigma 1/(n!)# from n is #[0,oo)# for convergence?

1 Answer
Jan 14, 2017

Use the ratio test to show the series' convergence.

Explanation:

We will use the ratio test. The ratio test says that the for the series #suma_n#, we can make a determination about its convergence by taking #L=lim_(ararroo)abs(a_(n+1)/a_n)#. Examine the value of #L#:

  • If #L>1#, then #suma_n# is divergent.
  • If #L=1#, then the test is inconclusive.
  • If #L<1#, then #suma_n# is (absolutely) convergent.

So for the series #sum_(n=0)^oo1/(n!)# we let #a_n=1/(n!)#. Then we see that

#L=lim_(nrarroo)abs((1/((n+1)!))/(1/(n!)))=lim_(nrarroo)abs((n!)/((n+1)!))#

This takes recalling a little bit about factorial. The definition of factorial states that #(n+1)! =(n+1)(n!)#, similar to how #7! = 7*6!#. Thus:

#L=lim_(nrarroo)abs((n!)/((n+1)(n!)))=lim_(nrarroo)abs(1/(n+1))=0#

Since #L=0# and therefore #L<1#, we see that #suma_n=sum_(n=0)^oo1/(n!)# is convergent through the ratio test.