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

Question

Isaiah Allen · Accepted 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.

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.

Emilia Aguilar · Answer

undefined To determine if the series $ \sum \frac{1}{n!} $ converges or diverges using the comparison test, we compare it to a known convergent or divergent series. In this case, since $ n! $ grows much faster than $ n^k $ for any positive integer $ k $, we can compare it to the series $ \sum \frac{1}{n^2} $, which is a convergent p-series with $ p = 2 $.

Therefore, we can establish the following comparison:
$$ 0 < \frac{1}{n!} < \frac{1}{n^2} $$

Since the series $ \sum \frac{1}{n^2} $ converges, and $ \frac{1}{n!} $ is smaller than $ \frac{1}{n^2} $ for all $ n $, by the comparison test, the series $ \sum \frac{1}{n!} $ also converges.