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

Question

Isaiah Allen · Accepted Answer

Use the ratio test to show the series' convergence.

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#: 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.

Cameron Alexander · Answer

undefined To test the convergence of the series Sigma 1/(n!) from n is [0,oo), you can use the ratio test.

Apply the ratio test by taking the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term. If the limit is less than 1, the series converges absolutely. If it's greater than 1, the series diverges. If the limit is exactly 1, the test is inconclusive.

In this case, the series is Sigma 1/(n!), so the nth term is 1/(n!).

Taking the ratio of the (n+1)th term to the nth term:

((n+1)!) / (n!)

This simplifies to (n+1).

Taking the limit as n approaches infinity:

lim (n+1) as n approaches infinity = infinity

Since the limit is greater than 1, the series diverges.