Question

Determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !} $$

Short answer

The given series converges.

Step-by-step solution

Write down the given series

The given series is \[ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !} \]

Apply the Ratio Test

Taking the absolute value, apply the ratio test: \[ \lim_ {n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_ {n \rightarrow \infty} \left| \frac{(-1)^{n+1}/(n + 1)!}{(-1)^n/n!} \right| \]

Simplify the Ratio

Simplify the ratio by cancelling out terms and taking the limit: \[ \lim_ {n \rightarrow \infty} \left| \frac{(-1)^{n+1}/(n + 1)!}{(-1)^n/n!} \right| = \lim_ {n \rightarrow \infty} \left| \frac{1}{n + 1} \right| = 0 \]

Determine Convergence/Divergence

Since the limit of the ratio is less than 1, it can be concluded that the given series converges according to the Ratio Test.