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

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 $n \ge 4 , n \in \mathbb{N}$. Let's show that it also holds for $n + 1 > 4$.

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

$\ge 4 \cdot {n}^{2}$, because $n + 1 > 4$ by hypothesis

$\ge {n}^{2} \left(1 + \frac{2}{n} + \frac{1}{n} ^ 2\right)$, because ${n}^{2} \ge n > 1$ by hypothesis

$= {n}^{2} + 2 n + 1 = {\left(n + 1\right)}^{2}$.

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