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Chapter 9: Problem 28

Use the Comparison Test or Limit Comparison Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{k^{2}+k-1}{k^{4}+4 k^{2}-3}$$

Short Answer

Expert verified
Answer: The series converges.

Step by step solution

01

Simplify the approximated series

Simplify the terms in the series by dividing the numerator by the denominator: $$\sum_{k=1}^{\infty} \frac{k^{2}}{k^{4}} = \sum_{k=1}^{\infty} \frac{1}{k^{2}}$$ This is a famous series called the p-series, where p=2.
02

Check convergence criteria for p-series

A p-series converges if p>1. In our case, p=2 which is greater than 1, so the p-series converges. Now, let's perform the Limit Comparison Test by taking the limit of the ratio of the series to its approximation.
03

Perform the Limit Comparison Test

Take the limit of the ratio between the given series and the approximated one as k approaches infinity: $$\lim_{k \to \infty} \frac{\frac{k^{2}+k-1}{k^{4}+4 k^{2}-3}}{\frac{1}{k^{2}}} = \lim_{k \to \infty} \frac{k^4(k^2+k-1)}{k^2(k^4+4k^2-3)}$$ Simplify the limit by canceling out terms: $$\lim_{k \to \infty} \frac{k^2+k-1}{k^4+4k^2-3}$$ We can use L'Hopital's Rule since both the numerator and denominator are approaching infinity: $$\lim_{k \to \infty} \frac{2k+1}{4k^3+8k} = \lim_{k \to \infty} \frac{1}{2k^2}$$ The limit exists and is equal to 0 as k approaches infinity.
04

Conclusion about convergence

Since the ratio of the given series and the approximated one goes to zero (a finite and positive value), by the Limit Comparison Test, the convergence of the given series must be the same as the convergence of the approximated series. Since the p-series converges (p=2), the given series also converges: $$\sum_{k=1}^{\infty} \frac{k^{2}+k-1}{k^{4}+4 k^{2}-3}$$ converges.

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Most popular questions from this chapter

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}$$

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty} \frac{(-1)^{k} \tan ^{-1} k}{k^{3}}$$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the value of the limit or determine that the limit does not exist. $$a_{n+1}=\sqrt{2+a_{n}} ; a_{0}=1, n=0,1,2, \dots$$

The Riemann zeta function is the subject of extensive research and is associated with several renowned unsolved problems. It is defined by \(\zeta(x)=\sum_{k=1}^{\infty} \frac{1}{k^{x}}\). When \(x\) is a real number, the zeta function becomes a \(p\) -series. For even positive integers \(p,\) the value of \(\zeta(p)\) is known exactly. For example, $$ \sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6}, \quad \sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}, \quad \text { and } \quad \sum_{k=1}^{\infty} \frac{1}{k^{6}}=\frac{\pi^{6}}{945}, \ldots $$ Use estimation techniques to approximate \(\zeta(3)\) and \(\zeta(5)\) (whose values are not known exactly) with a remainder less than \(10^{-3}\).

Consider the series \(\sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{p}},\) where \(p\) is a real number. a. For what values of \(p\) does this series converge? b. Which of the following series converges faster? Explain. $$ \sum_{k=2}^{\infty} \frac{1}{k(\ln k)^{2}} \text { or } \sum_{k=3}^{\infty} \frac{1}{k \ln k(\ln \ln k)^{2}} ? $$
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