Chapter 5: Problem 369
Let \(0
Short Answer
Expert verified
The series converges for \( p \leq -1 \).
Step by step solution
01
Understand the series components
The given series is \( \, \sum_{n=1}^{\infty} n^{p} r^{n} \, \), where \( 0 < r < 1 \). The terms of this series are \( n^p r^n \), which combines a polynomial term \( n^p \) with an exponential decay \( r^n \).
02
Consider the nature of convergence
For the series \( \sum_{n=1}^{\infty} n^{p} r^{n} \) to converge, the terms \( n^p r^n \) must tend to zero as \( n \) approaches infinity. However, since the series also involves an exponential decay component \( r^n \), it suggests you could apply comparison tests or ratio tests.
03
Apply the Ratio Test
Using the Ratio Test, examine the ratio of successive terms: \( \frac{a_{n+1}}{a_n} = \frac{(n+1)^p r^{n+1}}{n^p r^n} = \frac{(n+1)^p}{n^p}r \). As \( n \to \infty \), \( \frac{(n+1)^p}{n^p} \sim (1 + \frac{1}{n})^p \approx 1 \). Thus, the ratio simplifies to \( r \). For convergence, this ratio must be \( < 1 \).
04
Dominance of the exponential decay
Since \( r^n \) is dominant over \( n^p \) for large \( n \), convergence depends on the balance of terms \( n^p \) and \( r^n \). For convergence, the decay from \( r^n \) must outpace any increase in \( n^p \). If \( p \leq -1 \), the polynomial does not grow too fast, allowing \( r^n \) to dominate.
05
Utilize the Root Test for further confirmation
Consider also using the Root Test to confirm: \( \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n \to \infty} n^{p/n} r = r \). Since \( p/n \to 0 \) as \( n \to \infty \), the limit is \( r < 1 \), confirming convergence for \( p \leq -1 \).
06
Conclusion on convergence
The series \( \sum_{n=1}^{\infty} n^p r^n \) converges for real numbers \( p \leq -1 \) since the exponential decay in \( r^n \) sufficiently outpaces the polynomial growth of \( n^p \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ratio Test
The Ratio Test is a powerful tool to determine the convergence of infinite series. It involves examining the ratio of successive terms of the series. For a series \( \sum a_n \), the Ratio Test focuses on \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If this limit is less than 1, the series converges. If it is greater than 1 or infinite, the series diverges. If it equals 1, the test is inconclusive.
In our exercise involving the series \( \sum_{n=1}^{\infty} n^p r^n \), the ratio of successive terms can be simplified to the expression \( r \), due to the nature of \((n+1)^p/n^p \approx 1\) as \(n\) becomes large. Since \( r < 1 \), the test indicates convergence, confirming the dominance of the exponential decay term.
In our exercise involving the series \( \sum_{n=1}^{\infty} n^p r^n \), the ratio of successive terms can be simplified to the expression \( r \), due to the nature of \((n+1)^p/n^p \approx 1\) as \(n\) becomes large. Since \( r < 1 \), the test indicates convergence, confirming the dominance of the exponential decay term.
Root Test
The Root Test is another effective method to check for series convergence. It uses the \(n\)-th root of the terms in the series. To apply it, compute \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \). If this limit is less than 1, the series converges. If it is greater than 1, the series diverges. When this limit equals 1, the test does not provide a conclusion.
Applying the Root Test to \( \sum_{n=1}^{\infty} n^p r^n \), the \(n\)-th root of the term \(n^p r^n\) becomes \( n^{p/n} r \). As \( n \) approaches infinity, \( n^{p/n} \) tends to 1, leaving us with the term \( r \). Since \( r < 1 \), this confirms the series converges for \(p \leq -1\). The decay imposed by \(r^n\) ensures the series remains bounded.
Applying the Root Test to \( \sum_{n=1}^{\infty} n^p r^n \), the \(n\)-th root of the term \(n^p r^n\) becomes \( n^{p/n} r \). As \( n \) approaches infinity, \( n^{p/n} \) tends to 1, leaving us with the term \( r \). Since \( r < 1 \), this confirms the series converges for \(p \leq -1\). The decay imposed by \(r^n\) ensures the series remains bounded.
Exponential decay
Exponential decay refers to the process where quantities decrease rapidly at a rate proportional to their current value. In mathematical terms, this is often represented as \( r^n \) where \( 0 < r < 1 \). Such decay means that as \(n\) increases, \(r^n\) diminishes toward zero very quickly.
In context with the exercise, exponential decay is a critical factor in confirming convergence of the series \( \sum_{n=1}^{\infty} n^p r^n \). Since \(r^n\) becomes negligible more quickly than \(n^p\) can grow, it ensures the overall terms \(n^p r^n\) tend to zero, provided polynomial growth does not dominate. This strong decaying behavior greatly influences the convergence analysis, establishing it for \(p \leq -1\).
In context with the exercise, exponential decay is a critical factor in confirming convergence of the series \( \sum_{n=1}^{\infty} n^p r^n \). Since \(r^n\) becomes negligible more quickly than \(n^p\) can grow, it ensures the overall terms \(n^p r^n\) tend to zero, provided polynomial growth does not dominate. This strong decaying behavior greatly influences the convergence analysis, establishing it for \(p \leq -1\).
Polynomial growth
Polynomial growth refers to the manner in which a quantity increases in proportion to a power of its variable, often seen as \(n^p\). A polynomial term, like \(n^p\), grows as \(n\) becomes larger, but it does so at a slower rate compared to exponential functions.
In the given series, the term \(n^p\) competes against \(r^n\). When \(p\) is large, \(n^p\) grows substantially, potentially challenging the decaying effect of \(r^n\). However, if \(p \leq -1\), as discovered, \(n^p\) does not grow quickly enough to offset the rapid decay from \(r^n\). This consideration reveals why the series converges primarily for \(p \leq -1\), as exponential decay overpowers polynomial growth in these cases.
In the given series, the term \(n^p\) competes against \(r^n\). When \(p\) is large, \(n^p\) grows substantially, potentially challenging the decaying effect of \(r^n\). However, if \(p \leq -1\), as discovered, \(n^p\) does not grow quickly enough to offset the rapid decay from \(r^n\). This consideration reveals why the series converges primarily for \(p \leq -1\), as exponential decay overpowers polynomial growth in these cases.