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Chapter 5: Problem 224

For which \(p\) does the series \(\sum_{n=1}^{\infty} 2^{p n} / 3^{n}\) converge?

Short Answer

Expert verified
The series converges for \(p < \log_2(3)\).

Step by step solution

01

Identify the Series

The given series is \(\sum_{n=1}^{\infty} \frac{2^{pn}}{3^n}\). This is a series with terms of the form \(a_n = \frac{2^{pn}}{3^n}\).
02

Reformulating the Series

We can rewrite the terms of the series in terms of powers by observing that \(\frac{2^{pn}}{3^n} = \left( \frac{2^p}{3} \right)^n\). Thus, the series becomes \(\sum_{n=1}^{\infty} \left( \frac{2^p}{3} \right)^n\).
03

Recognize as a Geometric Series

The series \(\sum_{n=1}^{\infty} r^n\), where \(r = \frac{2^p}{3}\), is a geometric series. A geometric series converges if and only if the absolute value of the common ratio \(|r| < 1\).
04

Set the Convergence Condition

Since we want \(|r| < 1\), we need the absolute value of \(\frac{2^p}{3}\) to be less than 1: \(|\frac{2^p}{3}| < 1\).
05

Solve the Inequality

Solve the inequality \(|\frac{2^p}{3}| < 1\). This implies \(\frac{2^p}{3} < 1\), which translates to \(2^p < 3\). Taking the logarithm base 2 on both sides gives us \(p < \log_2(3)\).
06

Conclude the Solution

Thus, the series converges for \(p < \log_2(3)\). Using a calculator or tables, the approximate value of \(\log_2(3)\) is about 1.585.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Series Convergence
Converging series are essentially series that reach a specific limit as more terms are added. To determine if a series converges, especially a geometric one like \[ \sum_{n=1}^{\infty} r^n, \] we look at the value of the common ratio, often denoted as \( r \). The series will converge if and only if the absolute value of \( r \) is less than 1, meaning \[ |r| < 1. \]So, for our exercise, we established that the common ratio of the series is \( \frac{2^p}{3} \). We need this ratio's absolute value to be less than 1 for the series to converge.
  • Mathematically, a convergent series indicates a situation where ongoing additions of more terms will not change the sum beyond a certain point.
  • It remains finite and stable rather than growing indefinitely or oscillating without settling down.
Remember, such rules help us easily deal with infinite series in mathematics.
Geometric Series Ratio
The ratio used in a geometric series is a pivotal element for deciding series convergence. In each term of a geometric series, there is one constant factor that the previous term is multiplied by. This constant is called the common ratio, \( r \).For instance, in our given series, we found that the common ratio is\[ r = \frac{2^p}{3}. \]The properties of geometric series help us decide convergence based on the value of \( r \). These include:
  • The series converges if \( |r| < 1 \) and diverges if \( |r| \geq 1 \).
  • If the series converges, it approaches a finite sum \( S \) which can be calculated using the formula\[ S = \frac{1}{1-r}, \] assuming the first term of the series is 1.
Understanding these properties allows us to manipulate and analyze infinite series with ease.
Logarithm Properties
Logarithms are a mathematical tool used for solving problems involving exponential expressions, often simplifying calculations. In our exercise, logarithms help in determining for which values \( p \) the given series converges.The steps entailed taking the inequality\[ \frac{2^p}{3} < 1, \]and then manipulating it using logarithms to solve for \( p \). By taking the logarithm base 2 of both sides, we find:\[ p < \log_2(3). \]Key properties involved:
  • Logarithms can transform multiplicative relationships into additive ones, which are often easier to manage.
  • The change of base formula and various logarithm rules like \( \log_a(xy) = \log_a(x) + \log_a(y) \) assist in breaking down complex problems.
Utilizing these properties, we effectively solved for \( p \), confirming convergence when \( p \) is less than approximately 1.585, as found using logarithmic tables or a calculator.

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Most popular questions from this chapter

Suppose that \(\left|\frac{a_{n+1}}{a_{n}}\right| \leq(n+1)^{p}\) for all \(n=1,2, \ldots\) where \(p\) is a fixed real number. For which values of \(p\) is \(\sum_{n=1}^{\infty} n ! a_{n}\) guaranteed to converge?

In the following exercises, use an appropriate test to determine whether the series converges. $$a_{k}=2^{k} /\left(\begin{array}{l}3 k \\ k\end{array}\right)$$

State whether each of the following series converges absolutely, conditionally, or not at all. $$\sum_{n=1}^{\infty}(-1)^{n+1} n^{1 / n}$$

Suppose that \(\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|=p\). For which values of \(r>0\) is \(\sum_{n=1}^{\infty} r^{n} a_{n}\) guaranteed to converge?

Let \(a_{n}=2^{-[n / 2]}\) where \([x]\) is the greatest integer less than or equal to \(x\). Determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges and justify your answer. The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges.
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