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:

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

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