Question

Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n+1} n !}{1 \cdot 3 \cdot 5 \cdot \cdots(2 n+1)} $$

Short answer

The series is convergent.

Step-by-step solution

Define the series and the Ratio Test

The series in the task is \( \sum_{n=0}^{\infty} \frac{(-1)^{n+1} n !}{1 \cdot 3 \cdot 5 \cdot \cdots (2n+1)} \). The Ratio Test states that for a series \( \sum_{n=1}^{\infty} u_n \), it is convergent if \( \lim_{n \to \infty} \left|\dfrac{u_{n+1}}{u_n}\right| < 1 \), and divergent if the limit exceeds 1.

Calculate the ratio of consecutive terms

The ratio of the terms \( (n+1) \) and \( n \) is: \[\left|\frac{u_{n+1}}{u_n}\right| = \left|\frac{(-1)^{(n+1)+1} (n+1) !}{1 \cdot 3 \cdot 5 \cdot \cdots (2(n+1)+1)} \cdot \frac{1 \cdot 3 \cdot 5 \cdot \cdots (2n+1)}{(-1)^{n+1} n!}\right|= \left|\frac{(-1)(n+1)n!}{2(n+1)+1}\right|\]= \(\left|\frac{-(n+1)}{2n + 3}\right|\)

Determine the limit as n approaches infinity

The value of the limit as \( n \rightarrow \infty \) is:\[\lim_{n \rightarrow \infty} \left|\frac{-(n+1)}{2n + 3}\right| = \lim_{n \rightarrow \infty} \left|- \frac{1}{2} \right| = \frac{1}{2}\]The absolute value of the limit is less than 1.

Conclude the test

Since the absolute value of the limit is less than 1, according to the Ratio Test, the given series is convergent.