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Chapter 8: Problem 22

Show that the series diverges. \(\sum_{n=1}^{\infty} \frac{n}{\sqrt{2 n^{2}+1}}\)

Short Answer

Expert verified
We showed that the original series \( \sum_{n=1}^{\infty} \frac{n}{\sqrt{2n^{2}+1}} \) and the simpler divergent p-series \( \frac{1}{\sqrt{2n}} \) have the same behavior using the limit comparison test. The limit of the ratio of the given series to the simpler series as n tends to infinity equals to \( \frac{1}{\sqrt{2}} \neq 0 \). Thus, since the simpler series diverges, the original series also diverges.

Step by step solution

01

Identify a Simpler Series

A simpler series to compare with the original series \(\sum_{n=1}^{\infty} \frac{n}{\sqrt{2n^{2}+1}}\) is \(\sum_{n=1}^{\infty} \frac{n}{\sqrt{2n^2}}\), or \( \frac{1}{\sqrt{2n}}\). This is a p-series with \(p = 0.5\). According to the p-series convergence test, if \( p \leq 1 \), then the p-series diverges.
02

Apply the Limit Comparison Test

The limit comparison test checks the limit of the ratio of the original series to the simpler series as \( n \) tends to infinity. Here, the series are \( \frac{n}{\sqrt{2n^{2}+1}} \) and \( \frac{1}{\sqrt{2n}} \). Compute the limit as follows: \[ \lim_{{n \to \infty}} \frac{n / \sqrt{2n^{2}+1}}{1 / \sqrt{2n}} = \lim_{n \to \infty} \frac{n^{2}}{ \sqrt{2n^{2}+1}} \] Multiply the numerator and the denominator of the fraction inside the limit by \(n^{-2}\) to simplify it: \[ \lim_{{n \to \infty}} \frac{n^{2} \cdot n^{-2}}{ \sqrt{2n^{2}+1} \cdot n^{-2}} = \lim_{n \to \infty} \frac{1}{ \sqrt{2 + \frac{1}{n^{2}}}} \]
03

Compute the Limit

Now, compute the limit as \( n \) tends to infinity: \[ \lim_{n \to \infty} \frac{1}{ \sqrt{2 + \frac{1}{n^{2}}}} = \frac{1}{\sqrt{2 + 0}} = \frac{1}{\sqrt{2}} \neq 0 \] The result is not equal to 0, which means that our given series behaves the same way as the simpler series.
04

Conclude if the Series Converges or Diverges

Since we showed that the simpler series \( \frac{1}{\sqrt{2n}}\) diverges and the original series behaves the same way as the simpler series, it follows from the limit comparison test that the original series \( \sum_{n=1}^{\infty} \frac{n}{\sqrt{2n^{2}+1}} \) also diverges.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a valuable tool in determining whether a series converges or diverges. This test involves comparing a complex series with a simpler one that you already know about. For this to work, each series must have positive terms. You calculate a limit that compares the terms of the two series.
  • If the limit is a positive finite number, the series behave similarly. In other words, either both series converge or both diverge.
  • If the limit is zero and the simpler series converges, then the original series also converges.
  • If the limit is infinity and the simpler series diverges, then the original series diverges too.
To apply this test, find a comparable series with known behavior and check the limit.
P-Series
P-series are special mathematical series of the form \[\sum_{n=1}^{\infty} \frac{1}{n^p}\] where \( p \) is a constant. The convergence or divergence of the series depends on the value of \( p \)}.
  • If \( p > 1 \), the series converges.
  • If \( p \leq 1 \), the series diverges.
This makes p-series a great baseline for comparison when using tests like the Limit Comparison Test. They're simple to recognize and understand.
Convergence Tests
Convergence tests are strategies or criteria used to determine whether series converge (approach a certain number) or diverge (do not approach any number). There are several key convergence tests, each suited for particular situations:
  • Limit Comparison Test: Useful for comparing the behavior of two series based on the limit of their term ratios.
  • Direct Comparison Test: Compares terms directly to decide convergence or divergence.
  • Ratio Test: Examines the ratio of consecutive terms. Often used for series with factorials or exponential terms.
  • Root Test: Similar to the Ratio Test but uses the nth root of the terms.
Selecting the right test depends on the form and behavior of the series you're working with.
Mathematical Proof
Mathematical proofs are structured arguments that establish the validity of mathematical statements or assertions. In the context of series, proofs help confirm whether a series converges or diverges.
Steps to a good proof:
  • Identify: Recognize the type of series or problem.
  • Select the Appropriate Test: Use the convergence tests to plan your approach.
  • Apply the Test: Work through the test methodically.
  • Conclude: Draw a clear, logical conclusion based on your work.
A thorough proof leaves no room for doubt, logically leading from assumptions to conclusions based on clear, defined rules.

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Most popular questions from this chapter

Force Exerted by a Charge Distribution Suppose that a charge \(Q\) is distributed uniformly along the positive \(x\) -axis from \(x=0\) to \(x=a\) and that a negative charge \(-Q\) is distributed uniformly along the negative \(x\) -axis from \(x=-a\) to \(x=0 .\) If a positive charge \(q\) is placed on the positive \(x\) -axis a distance of \(x\) units \((x>a)\) from the origin, then the force exerted by the charge distribution on \(q\) has magnitude $$ F=\frac{q Q}{4 \pi \varepsilon_{0} a}\left[\frac{1}{x-a}+\frac{1}{x+a}-\frac{2}{x}\right] $$ and direction along the positive \(x\) -axis. Show that if \(x\) is large, then $$ F \approx \frac{q Q a}{2 \pi \varepsilon_{0} x^{3}} $$

Use the binomial series to find the power series representation of the function. Then find the radius of convergence of the series. \(f(x)=\sqrt{1-x^{2}}\)

Use the Maclaurin series for \(\cos x\) to calculate \(\cos 3^{\circ}\) accurate to five decimal places.

Use Equation (1) to find the Taylor series of \(f\) at the given value of \(c .\) Then find the radius of convergence of the series. \(f(x)=\sin x, \quad c=\frac{\pi}{4}\)

Find all values of \(x\) for which the series \(\sum_{n=1}^{\infty} \frac{x^{n}}{n}\) (a) converges absolutely and (b) converges conditionally.
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