Question

Test the following series for convergence. $$\sum_{n=1}^{\infty} \frac{(-3)^{n}}{n !}$$

Short answer

The series converges.

Step-by-step solution

Identify the Series

The given series to test for convergence is \[ \sum_{n=1}^{\infty} \frac{(-3)^{n}}{n !} \]. This is an infinite series.

Recognize the Type of Series

Notice that this series resembles the form of a power series or exponential series of the general form \( \sum_{n=0}^{\infty} \frac{x^{n}}{n !} \).

Apply the Ratio Test

To test for convergence, use the ratio test. The ratio test involves finding the limit \( L \) \as follows:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

Calculate the Ratio

Substitute the terms of the series:\[a = \frac{(-3)^{n}}{n!} \]\[ \frac{a_{n+1}}{a_{n}} = \frac{\left(\frac{(-3)^{n+1}}{(n+1) !} \right)}{\left(\frac{(-3)^{n}}{n !} \right)} \]\[ = \frac{(-3)^{n+1}}{(n+1) !} \times \frac{n !}{(-3)^{n}} \]\[ = \frac{-3 \cdot (-3)^{n}}{(n+1) \cdot n !} \times \frac{n !}{(-3)^{n}} \]\[ = \frac{-3}{n+1} \]

Take the Limit

Now, compute the limit:\[ L = \lim_{n \to \infty} \left| \frac{-3}{n+1} \right| = \lim_{n \to \infty} \frac{3}{n+1} = 0 \]

Interpret the Result

Since \(L = 0 \), and \(0 < 1 \), by the ratio test the series \( \sum_{n=1}^{ \infty} \frac{(-3)^{n}}{n !} \) converges.

Key concepts

ratio test

The ratio test is a method we use to determine if an infinite series converges. It is particularly helpful for series where terms include factorials or exponential functions. To apply the ratio test, we find the limit of the absolute value of the ratio of consecutive terms. Here's the formula:

\text{\[ L = \lim_{{n \to \infty}} \left| \frac{a_{{n+1}}}{a_n} \right| \]}
If this limit, \( L \), is less than 1, the series converges. If \( L \) is greater than 1, the series diverges. And if \( L \) equals 1, the test is inconclusive, and we might need to use another test to determine convergence.

infinite series

An infinite series is the sum of an infinite sequence of terms. Formally, an infinite series is written as: \[ \sum_{{n = 1}}^{{\infty}} a_n \]
Studying whether an infinite series converges or diverges means determining if the sum approaches a finite value as more terms are added. There are various tests to determine convergence, including:

• The Ratio Test
• The Root Test
• The Comparison Test
• The Integral Test

For example, the series \[ \sum_{{n=1}}^{{\infty}} \frac{(-3)^{n}}{n!} \]
is the given infinite series in the exercise, and we determined its convergence using the ratio test in the solution.

exponential series

An exponential series involves terms that include expressions of the form \( x^n \) divided by a factorial \( n! \). These series are related to the exponential function: \[ e^x = \sum_{{n=0}}^{{\infty}} \frac{x^n}{n!} \]
This series converges for all real numbers \( x \). If we recognize a given series resembling this form, it's fairly straightforward to analyze its convergence using similar techniques.

In our example from the exercise, the series \[ \sum_{{n=1}}^{{\infty}} \frac{(-3)^{n}}{n!} \]
resembles an exponential series with \( x = -3 \). By applying the ratio test, we confirm the series converges, similar to the exponential series \( e^x \) converging for all \( x \).