How do you use the direct comparison test to determine if #Sigma 1/(n!)# from #[0,oo)# is convergent or divergent?

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Answer 1 · 2017-03-19T11:57:32

#sum_(n=0) 1/(n!) = e#

We have:

#a_n = 1/(n!) = 1/(1*2*3*...*n) < 1/(2*2*...*2) = (1/2)^(n-1)#

The series:

#sum_(n=0)^oo (1/2)^(n-1)#

can be expressed as the sum of a geometric series of ratio #1/2# and as #1/2 < 1# such series is convergent:

#sum_(n=0)^oo (1/2)^(n-1)= 2 sum_(n=0)^oo (1/2)^n = 2/(1-1/2) =4#

By direct comparison we can then conclude that:

#sum_(n=0) 1/(n!)#

is also convergent, ant that its sum is less than #4#.
In fact if we recall that the MacLaurin series of the exponential function is:

#e^x = sum_(n=0)^oo x^n/(n!)#

we can see that:

#sum_(n=0) 1/(n!) = e^1 = e#