How do you determine whether #1/(n!)# convergence or divergence with direct comparison test?

To determine the convergence or divergence of the series ( \sum \frac{1}{n!} ) using the direct comparison test, you can compare it to a known series whose convergence behavior is already established.

The series ( \sum \frac{1}{n!} ) is known as the Taylor series expansion for the exponential function ( e^x ) evaluated at ( x = 1 ): [ e = \sum_{n=0}^{\infty} \frac{1}{n!} ]

Given that the series for ( e ) is convergent, we can use it as a comparison series.

Using the direct comparison test: If ( 0 \leq \frac{1}{n!} \leq \frac{1}{n} ) for all ( n ), then ( \sum \frac{1}{n!} ) is convergent.

Since ( \frac{1}{n!} ) is always smaller than ( \frac{1}{n} ) for ( n \geq 2 ) (excluding ( n = 1 )), and the series ( \sum \frac{1}{n} ) is a p-series with ( p = 1 ) (which is known to be divergent), we can conclude that ( \sum \frac{1}{n!} ) is convergent by comparison to the convergent series for ( e ).