Question

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

Answer 1

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.