Question

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

Short answer

The given series converges.

Step-by-step solution

Recognize the Type of Series

The given series is an alternating series, where the general term can be expressed as \(a_{n}=\frac{(-1)^{n} 2^{n}}{n !}\).

Calculate Ratio

The ratio of consecutive terms is \(\frac{a_{n+1}}{a_{n}} = \frac{(-1)^{n+1} 2^{n+1}}{(n+1) !} ÷ \frac{(-1)^{n} 2^{n}}{n !} = \frac{(-1) 2}{n+1}\). We take the absolute value to get \(|\frac{a_{n+1}}{a_{n}}| = \frac{2}{n+1}\)

Apply Ratio Test

According to the Ratio Test, we now calculate the limit \(\lim_{n \to \infty} \frac{2}{n+1}\). This limit is 0 which is less than 1.

Interpreting the Result

As per the Ratio Test, if the limit of the ratio is less than 1, the series is absolutely convergent. Therefore, the given series is absolutely convergent.