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Chapter 9: Problem 33

Verify that the Ratio Test is inconclusive for the \(p\) -series. $$ \sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}} $$

Short Answer

Expert verified
The ratio test is inconclusive for the given p-series. The limit of the ratio of each term to the previous term is equal to 1 when n goes to infinity.

Step by step solution

01

Expression of sequence ratio

First, express the ratio of the (n+1)-th term to the nth term of the sequence. This gives:\( \frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)^{3/2}}}{\frac{1}{n^{3/2}}} = \left( \frac{n}{n+1} \right)^{3/2} \)
02

Limit calculation

Next, calculate the limit of the ratio as n approaches infinity. This yields:\( \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^{3/2} = \lim_{n \to \infty} (1 - \frac{1}{n+1})^{3/2} \)
03

Result and interpretation

When evaluated, the final limit simplifies down to 1. According to the ratio test, when the limit equals one, the test is inconclusive. Therefore, the ratio test proves to be inconclusive for the given p-series.

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Key Concepts

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

p-series
A p-series is a type of infinite series expressed in the form \[ \sum_{n=1}^{\infty} \frac{1}{n^p}, \] where \( p \) is a positive constant. This kind of series is of particular interest in the study of convergence and divergence. The behavior of a p-series depends on the value of \( p \):
  • If \( p > 1 \), the p-series is known to converge.
  • If \( p \leq 1 \), the p-series diverges.
This exercise involves a p-series with \( p = \frac{3}{2} \), thus you might expect it to converge since \( \frac{3}{2} > 1 \). This makes p-series a vital component in understanding various convergence tests, including the Ratio Test.
limit calculation
Limit calculation is crucial for applying the Ratio Test, a tool used for determining the convergence of an infinite series. In the context of this exercise:You express the ratio of the consecutive terms of the series:\[ \frac{a_{n+1}}{a_n} = \left( \frac{n}{n+1} \right)^{3/2}. \]Then, you calculate the limit of this ratio as \( n \to \infty \) to determine the series' behavior:\[ \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^{3/2}. \]As \( n \) becomes very large, \( \frac{n}{n+1} \approx 1 \), and thus\[ \lim_{n \to \infty} \left( 1-\frac{1}{n+1} \right)^{3/2} = 1. \]This calculation shows that the limit approaches 1, indicating a special situation when using the Ratio Test.
convergent series
A convergent series is one where the sum of its infinite terms approaches a finite number. This property allows for predicting behavior over an infinite range, which is valuable in mathematical analysis and applications.For this p-series:Since \( p = \frac{3}{2} \) is greater than 1, mathematical theory suggests the series should converge. However, verification using convergence tests such as the Ratio Test is typically conducted:
  • Confirm predictions by examining specific cases.
  • Refine understanding by analyzing associated limits.
Thus, although theoretical evidence suggests convergence for this series, tests are still important for practical confirmation.
inconclusive test
When performing the Ratio Test on the p-series provided, we reach an interesting point. The Ratio Test requires us to find the limit of the ratio of successive terms. However:
  • The calculated limit is \( 1 \).
  • According to the Ratio Test criteria, if this limit equals 1, the test is deemed inconclusive.
This result means that the Ratio Test does not provide us with a conclusive answer regarding whether the series converges or diverges. It highlights the limitation of the Ratio Test:While the Ratio Test is effective in many scenarios (when the limits are clearly less than 1 or greater than 1), cases where the limit equals 1 require alternative methods. Understanding why a test might be inconclusive can deepen comprehension and guide more tailored approaches for specific series.

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Most popular questions from this chapter

Prove that if \(\left\\{s_{n}\right\\}\) converges to \(L\) and \(L>0\), then there exists ? number \(N\) such that \(s_{n}>0\) for \(n>N\).

Multiplier Effect The annual spending by tourists in a resort city is \(\$ 100\) million. Approximately \(75 \%\) of that revenue is again spent in the resort city, and of that amount approximately \(75 \%\) is again spent in the same city, and so on. Write the geometric series that gives the total amount of spending generated by the \(\$ 100\) million and find the sum of the series.

Find the Maclaurin series for \(f(x)=\ln \frac{1+x}{1-x}\) and determine its radius of convergence. Use the first four terms of the series to approximate \(\ln 3 .\)

Investigation The interval of convergence of the series \(\sum_{n=0}^{\infty}(3 x)^{n}\) is \(\left(-\frac{1}{3}, \frac{1}{3}\right)\) (a) Find the sum of the series when \(x=\frac{1}{6}\). Use a graphing utility to graph the first six terms of the sequence of partial sums and the horizontal line representing the sum of the series. (b) Repeat part (a) for \(x=-\frac{1}{6}\). (c) Write a short paragraph comparing the rate of convergence of the partial sums with the sum of the series in parts (a) and (b). How do the plots of the partial sums differ as they converge toward the sum of the series? (d) Given any positive real number \(M\), there exists a positive integer \(N\) such that the partial sum \(\sum_{n=0}^{N}\left(3 \cdot \frac{2}{3}\right)^{n}>M\) Use a graphing utility to complete the table. $$ \begin{array}{|l|l|l|l|l|} \hline \boldsymbol{M} & 10 & 100 & 1000 & 10,000 \\ \hline \boldsymbol{N} & & & & \\ \hline \end{array} $$

Suppose that \(\sum a_{n}\) diverges and \(c\) is a nonzero constant. Prove that \(\Sigma c a_{n}\) diverges.
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