Question

Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{1}{n !} $$

Short answer

The series \(\sum_{n=0}^{\infty} \frac{1}{n !}\) converges.

Step-by-step solution

Identify Known Series

In order to apply the Direct Comparison Test, a known series needs to be identified that can be compared to the given series. An applicable known series is the geometric series of \(\sum_{n=0}^{\infty} \frac{1}{2^n}\), which converges because the common ratio (1/2) is between -1 and 1.

Comparison

Now, compare the given series \(\sum_{n=0}^{\infty} \frac{1}{n !}\) with the known series \(\sum_{n=0}^{\infty} \frac{1}{2^n}\). For \(n\) greater than zero, \(\frac{1}{n !} <= \frac{1}{2^n}\), since \(n !\) is always less than or equal to \(2^n\). Therefore, the term \(a_n = \frac{1}{n !}\) of the given series is less than or equal to that of the known series \(b_n = \frac{1}{2^n}\).

Apply Direct Comparison Test

The Direct Comparison Test states that if \(0 <= a_n <= b_n\) for all \(n\) and \(\sum_{n=0}^{\infty} b_n\) converges, then \(\sum_{n=0}^{\infty} a_n\) also converges. Since \(\sum_{n=0}^{\infty} \frac{1}{2^n}\) is a convergent geometric series and \(0 <= \frac{1}{n !} <= \frac{1}{2^n}\), \(\sum_{n=0}^{\infty} \frac{1}{n !}\) also converges by the Direct Comparison Test.