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Chapter 8: Problem 8

Zeigen Sie, dass die folgenden Reihen absolut konvergieren: (a) \(\left(\sum_{n=1}^{\infty} \frac{2+(-1)^{n}}{2^{n-1}}\right)\) (b) \(\left(\sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n}\left(\frac{1}{3}+\frac{1}{n}\right)^{n}\right)\) (c) \(\left(\sum_{n=1}^{\infty}\left(\begin{array}{l}4 n \\ 3 n\end{array}\right)^{-1}\right)\)

Short Answer

Expert verified
Question: Determine whether the following series converge absolutely: a) \(\sum_{n=1}^{\infty} \frac{2+(-1)^{n}}{2^{n-1}}\) b) \(\sum_{n=1}^{\infty} (-1)^{n}\frac{1}{n}\left(\frac{1}{3}+\frac{1}{n}\right)^{n}\) c) \(\sum_{n=1}^{\infty} \left(\begin{array}{l}4 n \\\ 3 n\end{array}\right)^{-1}\)

Step by step solution

01

Series (a)

For series (a), let \(a_n = \frac{2+(-1)^{n}}{2^{n-1}}\). We must show that \(\sum_{n=1}^{\infty} {|a_n|}\) converges.
02

Applying the Comparison Test

Notice that for all \(n\), \(0 \le \frac{2+(-1)^{n}}{2^{n-1}} \le \frac{3}{ 2^{n-1}}\). We will compare \(|a_n|\) with \(\frac{3}{ 2^{n-1}}\) using the Comparison Test.
03

Convergence of the reference series

Since \(\sum_{n=1}^{\infty} \frac{3}{ 2^{n-1}}\) is a geometric series with a common ratio of \(\frac{1}{2}\), which is less than 1, it converges.
04

Applying the Comparison Test

By the Comparison Test, since \(\sum_{n=1}^{\infty} \frac{3}{ 2^{n-1}}\) converges and \(0 \le |a_n| \le \frac{3}{ 2^{n-1}}\) for all \(n\), the series \(\sum_{n=1}^{\infty} {|a_n|}\) also converges. Hence, the series (a) converges absolutely.
05

Series (b)

For series (b), let \(f_n = (-1)^{n}\frac{1}{n}\left(\frac{1}{3}+\frac{1}{n}\right)^{n}\). To prove the convergence of the absolute value of the series, we must show that \(\sum_{n=1}^{\infty} {|f_n|}\) converges.
06

Ratio Test

We want to apply the Ratio Test, so we need to calculate \(\lim_{n \to \infty} \frac{|f_{n+1}|}{|f_n|}\).
07

Computing the limit

Calculating the limit, \(\lim_{n \to \infty} \frac{|f_{n+1}|}{|f_n|} = \lim_{n \to \infty} \frac{(n)\left(\frac{1}{3}+\frac{1}{n}\right)^{n}}{(n+1)\left(\frac{1}{3}+\frac{1}{n+1}\right)^{n+1}}\). Applying L'H么pital's rule, we obtain \(\lim_{n \to \infty} \frac{\left(\frac{1}{3}+\frac{1}{n}\right)^{n}}{\left(\frac{1}{3}\right)^{n+1}}\). As \(n\) goes to infinity, the limit \(\lim_{n \to \infty} \frac{\left(\frac{1}{3}+\frac{1}{n}\right)^{n}}{\left(\frac{1}{3}\right)^{n+1}} = 0 < 1\).
08

Series Convergence

By the Ratio Test, the series \(\sum_{n=1}^{\infty} {|f_n|}\) converges absolutely. Hence, the series (b) converges absolutely.
09

Series (c)

For series (c), let \(g_n = \left(\begin{array}{l}4 n \\\ 3 n\end{array}\right)^{-1}\). To prove the absolute convergence, we need to show that \(\sum_{n=1}^{\infty} {|g_n|}\) converges.
10

Applying the Ratio Test

We want to apply the Ratio Test, so we need to calculate \(\lim_{n \to \infty} \frac{|g_{n+1}|}{|g_n|}\).
11

Computing the limit

Calculating the limit, \(\lim_{n \to \infty} \frac{|g_{n+1}|}{|g_n|} = \lim_{n \to \infty} \frac{\left(\begin{array}{l}4 (n+1) \\\ 3 (n+1)\end{array}\right)}{\left(\begin{array}{l}4 n \\\ 3 n\end{array}\right)} = \lim_{n \to \infty} \frac{4(n+1)}{4n} = 1\). Since the limit is equal to \(1\), the Ratio Test is inconclusive.
12

New approach: P-Test

However, we can rewrite the series as \(\sum_{n=1}^{\infty} \frac{1}{\left(\frac{4}{3}n\right)}\) which is similar to the form of \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). In our case, we have \(p = 1\). Since \(p = 1 \le 1\), by the P-Test, the series \(\sum_{n=1}^{\infty} {|g_n|}\) does not converge. Hence, the series (c) does not converge absolutely.

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