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Chapter 10: Problem 19

Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{3^{n}}{n !} $$

Short Answer

Expert verified
The series \( \sum_{n=0}^{\infty} \frac{3^{n}}{n !} \) converges according to the Ratio Test.

Step by step solution

01

Identify the series

The series given is \( \sum_{n=0}^{\infty} \frac{3^{n}}{n !} \). The Ratio Test involves comparing the ratio of consecutive terms, so the primary objective here is to formulate a ratio involving the (n+1)th and nth terms of this series.
02

Apply the Ratio Test

To apply the Ratio Test, one needs to compute the absolute value of the ratio of the (n+1)th term to the nth term, and compare this with 1. This ratio is given by \( \left|\frac{a_{n+1}}{a_n}\right| \), where \( a_n = \frac{3^{n}}{n !} \). Consider the ratio \( \left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\frac{3^{n+1}}{(n+1) !}}{\frac{3^{n}}{n !}}\right| = \frac{3}{n+1} \).
03

Test for convergence

Next, one determines whether our series converges or diverges. According to the Ratio Test, the series converges if the ratio \( \left|\frac{a_{n+1}}{a_n}\right| \) is less than 1 for large n, and diverges if the ratio is more than 1. In this case, since \( \frac{3}{n+1} \) goes to 0 as \( n \) goes to infinity, thus the ratio is less than 1, indicating that the series converges.

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