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Chapter 9: Problem 67

Neither the Ratio Test nor the Root Test helps with \(p\) -series. Try them on $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ and show that both tests fail to provide information about convergence.

Short Answer

Expert verified
Both the Ratio Test and the Root Test are inconclusive for the \( p \)-series.

Step by step solution

01

Applying the Ratio Test

To apply the Ratio Test, we need to consider the ratio of consecutive terms of the series. Let \( a_n = \frac{1}{n^p} \). The Ratio Test states that for the series \( \sum a_n \), we consider \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), the series converges; if \( L > 1 \), the series diverges; and if \( L = 1 \), the test is inconclusive.Calculate the ratio: \[\frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)^p}}{\frac{1}{n^p}} = \frac{n^p}{(n+1)^p} = \left( \frac{n}{n+1} \right)^p.\]Calculate the limit:\[L = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^p = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right)^p = 1^p = 1.\]Since \( L = 1 \), the Ratio Test is inconclusive.
02

Applying the Root Test

For the Root Test, we consider the expression \( L = \lim_{n \to \infty} \sqrt[n]{|a_n|} \). If \( L < 1 \), the series converges; if \( L > 1 \), the series diverges; and if \( L = 1 \), the test is inconclusive.Calculate the expression:\[L = \lim_{n \to \infty} \sqrt[n]{\frac{1}{n^p}} = \lim_{n \to \infty} \left( \frac{1}{n^p} \right)^{1/n} = \lim_{n \to \infty} n^{-p/n}.\]Evaluate the limit:\[L = \lim_{n \to \infty} n^{-p/n} = \lim_{n \to \infty} e^{-p \ln(n)/n} = e^0 = 1.\]Since \( L = 1 \), the Root Test is inconclusive as well.

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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 for determining the convergence of an infinite series. This test involves examining the limit of the absolute ratio of consecutive terms within a series. Essentially, you'd perform the following steps:
  • Identify the sequence of terms, denoted by \( a_n \), of the given series \( \sum a_n \).
  • Calculate the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \).
  • Compute the limit of this ratio as \( n \to \infty \), which results in \( L \).
  • Interpret the results in terms of convergence: if \( L < 1 \), the series converges; if \( L > 1 \), it diverges; if \( L = 1 \), the test is inconclusive.
For example, when applied to the \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), you find that \( L = 1 \). Under these circumstances, the Ratio Test yields no conclusion about the convergence of the series, rendering it inconclusive. The design of the Ratio Test works well for many types of series but not when \( L \) equals 1. This highlights an inherent limit in the test."
Root Test
The Root Test, like the Ratio Test, assesses the convergence of a series through a limit. Instead of using the ratio of consecutive terms, it examines the n-th root of individual terms:
  • Start with the term \( a_n \) from the series \( \sum a_n \).
  • Calculate \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \) to find \( L \).
  • Make the same convergence checks as with the Ratio Test: the series converges if \( L < 1 \), diverges if \( L > 1 \), and remains inconclusive if \( L = 1 \).
When using the Root Test on the \( p \)-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), the limit \( L \) also equals 1, leaving the test inconclusive. This similarity with the Ratio Test's outcome showcases the overlapping limits of these methodologies. Despite providing great precision in other contexts, the Root Test can also leave gaps in certain series assessments."
p-series
A \( p \)-series is a fundamental sequence form in the realm of infinite series, recognized by its general definition \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The convergence of such series depends largely on the value of \( p \):
  • If \( p > 1 \), the series converges. This conclusion leverages integral test results and other convergence theorems.
  • If \( p \leq 1 \), the series diverges. In these situations, the terms do not tend toward zero fast enough to sum to a finite limit.
Though neither the Ratio nor Root Tests directly settle convergence for \( p \)-series, understanding \( p \)-series based on \( p \) offers crucial insights into their behavior. They act as vital building blocks in analyzing more complex mathematical phenomena and underscore the need for various convergence tools, reinforcing that no single test is universally applicable."

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Most popular questions from this chapter

Series for \(\tan ^{-1} x\) for \(|x|>1\) Derive the series $$\begin{array}{l}\tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1 \\\\\tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1\end{array}$$ by integrating the series $$\frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots$$ in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x\).

a. Find the interval of convergence of the power series $$\sum_{n=0}^{\infty} \frac{8}{4^{n+2}} x^{n}$$ b. Represent the power series in part (a) as a power series about \(x=3\) and identify the interval of convergence of the new series. (Later in the chapter you will understand why the new interval of convergence does not necessarily include all of the numbers in the original interval of convergence.)

Which of the series \(\Sigma_{n=1}^{\infty} a_{n}\) defined by the formulas converge, and which diverge? Give reasons for your answers. $$a_{1}=3, \quad a_{n+1}=\frac{n}{n+1} a_{n}$$

The series $$e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\frac{x^{5}}{5 !}+\cdots$$ converges to \(e^{x}\) for all \(x\). a. Find a series for \((d / d x) e^{x} .\) Do you get the series for \(e^{x} ?\) Explain your answer. b. Find a series for \(\int e^{x} d x .\) Do you get the series for \(e^{x} ?\) Explain your answer. c. Replace \(x\) by \(-x\) in the series for \(e^{x}\) to find a series that converges to \(e^{-x}\) for all \(x\). Then multiply the series for \(e^{x}\) and \(e^{-x}\) to find the first six terms of a series for \(e^{-x} \cdot e^{x}\).

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=2}^{\infty} \frac{n}{(\ln n)^{(n / 2)}}$$
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