How do you apply the ratio test to determine if #Sigma 2^n/(n!)# from #n=[1,oo)# is convergent to divergent?

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Answer 1 · 2017-01-17T01:14:15

The series is convergent and:

#sum_(n=1)^oo 2^n/(n!) = e^2-1#

The ratio test states that a sufficient condition for a series:

#sum_(n=0)^oo a_n#

to converge absolutely is that:

#L = lim_(n->oo) abs(a_(n+1)/a_n) < 1#

We then evaluate the ratio for the series:

#sum_(n=1)^oo 2^n/(n!)#

#abs(a_(n+1)/a_n) = (2^(n+1)/((n+1)!))/(2^n/(n!)) = 2^(n+1)/2^n (n!)/((n+1)!) = 2/(n+1)#

so that:

#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) 2/(n+1) = 0#

proving that the series is convergent.

We should note by the way that the sum is:

#sum_(n=1)^oo 2^n/(n!) = -1 + [sum_(n=0)^oo x^n/(n!)]_(x=2) = e^2-1#