Question

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

Answer 1

The series is convergent and:

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

Explanation:

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`