Question

Determine whether the series converges conditionally or absolutely, or diverges. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1) !} $$

Short answer

The series both conditionally and absolutely converges.

Step-by-step solution

Identify the Series Test to Use

The series is an alternating series because of the \((-1)^n\) term. We will also need to consider the Absolute Convergence Test, so we will use both the Alternating Series Test and the Absolute Convergence Test.

Apply the Alternating Series Test

For the Alternating Series Test, the two conditions that need to be satisfied are: 1) The limit of the sequence must be zero and 2) The sequence must be decreasing. In this case, we have \(\lim_{n \to \infty} \frac{1}{(2n+1)!} = 0\) and \(\frac{1}{(2n+1)!} > \frac{1}{(2(n+1)+1)!}\) for all \(n\), so both conditions are satisfied, implying that the series converges conditionally.

Apply the Absolute Convergence Test

To apply the Absolute Convergence Test, we take the absolute value of the terms in the series, getting rid of the \(-1^n\) to have the series \(\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}\). Using the Ratio Test on this series, we see that \(\frac{\frac{1}{(2(n+1)+1)!}}{\frac{1}{(2n+1)!}} = \frac{(2n+1)!}{(2n+1+2)!} = 0 < 1\), so this series absolutely converges.