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Chapter 10: Problem 27

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

Expert verified
The series \(\sum_{n=0}^{\infty} \frac{(-1)^{n} 2^{n}}{n !}\) converges according to the Ratio Test.

Step by step solution

01

Write down the (n+1)th term and the nth term

Let's define \(a_{n+1}\) as the (n+1)th term and \(a_n\) as the nth term of the series. Here, \(a_{n+1} = \frac{(-1)^{n+1}2^{n+1}}{(n+1)!}\) and \(a_n = \frac{(-1)^n2^n}{n!}\).
02

Apply the Ratio Test

To apply the Ratio Test, we need to find the limit as n approaches infinity of the absolute value of the ratio of \(a_{n+1}\) to \(a_n\). This is expressed as \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
03

Simplify the ratio

Let's simplify \(\left| \frac{a_{n+1}}{a_n} \right|\) to get \(L = \lim_{n \to \infty} \left| \frac{(-1)^{n+1} 2^{n+1} n!}{(-1)^n 2^n (n+1)!} \right|\). The terms (-1)^(n+1)/(-1)^n simplify to -1, and 2^(n+1)/2^n simplifies to 2. Thus the expression becomes \(L = \lim_{n \to \infty} \left| \frac{-2 n!}{(n+1)!} \right|\). Given that (n+1)! = (n+1)n!, further simplification gives \(L = \lim_{n \to \infty} \left| \frac{-2}{n+1} \right|\).
04

Evaluate the limit

Taking the limit as n approaches infinity of the expression \(\left| \frac{-2}{n+1} \right|\) gives \(L = 0\), since as n goes to infinity, the expression goes to 0.
05

Determine convergence or divergence using the Ratio Test

The Ratio Test states that the series converges if \(L < 1\). In this case, \(L = 0 < 1\). Therefore, according to the Ratio Test, the series converges.

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