How do you determine if series 1/n!1n! converge or diverge with comparison test?

1 Answer
Feb 26, 2016

You first prove by induction that n! >= n^2 AA n>= 4, n in NN iff 0 <= 1/(n!) <= 1/n^2 AA n>=4, n in NN and you conclude by the comparison test that the series converges.

Explanation:

Basis : If n= 4, n! = 24 >= n^2 = 16.

Inductive step : Let's suppose n! >= n^2 is true for all n>=4, n in NN. Let's show that it also holds for n+1 > 4.

(n + 1)! = n! * (n+1) >= n^2 * (n + 1), by induction hypothesis

>= 4*n^2, because n+1 > 4 by hypothesis

>= n^2(1 + 2/n + 1/n^2), because n^2 >= n > 1 by hypothesis

= n^2 +2n + 1 = (n + 1)^2.

You can now conclude by mathematical induction that n! >= n^2 AA n>= 4, n in NN iff 0 <= 1/(n!) <= 1/n^2 AA n>=4, n in NN.

Since the series of general term 1/n^2 converges, you can conclude by the comparison test that the series of general term 1/(n!) also converges.