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

1 Answer
Feb 26, 2016

Answer:

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.