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Chapter 8: 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
Question: Determine the convergence of the series using the Comparison Test or Limit Comparison Test. $$\sum_{k=1}^{\infty} \frac{k^{2}+k-1}{k^{4}+4 k^{2}-3}$$ Answer: The given series converges by the Limit Comparison Test.

Step by step solution

01

Simplify the series and identify the comparison series

The given series is: $$\sum_{k=1}^{\infty} \frac{k^{2}+k-1}{k^{4}+4 k^{2}-3}$$ We can compare it to a simpler series formed by the highest powers of k in both numerator and denominator: $$\sum_{k=1}^{\infty} \frac{k^{2}}{k^{4}}$$
02

Perform the Limit Comparison Test

To apply the Limit Comparison Test, we will calculate the limit of the ratio of the given series and the comparison series, as k approaches infinity: $$\lim_{k \to \infty} \frac{\frac{k^{2}+k-1}{k^{4}+4 k^{2}-3}}{\frac{k^{2}}{k^{4}}}$$ This simplifies to: $$\lim_{k \to \infty} \frac{k^{2}(k^{2}+k-1)}{k^{4}(k^{4}+4 k^{2}-3)}$$
03

Simplify the limit and calculate

Now, simplify the limit expression: $$\lim_{k \to \infty} \frac{k^{4}+k^{3}-k^{2}}{k^{8}+4 k^{6}-3 k^{4}}$$ Divide both numerator and denominator by the highest power of k, which is \(k^{4}\): $$\lim_{k \to \infty} \frac{1+\frac{1}{k}-\frac{1}{k^{2}}}{k^{4}+4 \frac{1}{k}-3 \frac{1}{k^{2}}}$$ As k approaches infinity, the terms with k in the denominator go to zero: $$\lim_{k \to \infty} \frac{1}{k^{4}} = 0$$
04

Determine the convergence of the given series

Since we have compared our series with the series \(\sum_{k=1}^{\infty} \frac{k^{2}}{k^{4}}\), which is the same as \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}\) and is a convergent p-series (p=2 > 1), and the limit is 0, the given series converges as well. This is according to the Limit Comparison Test. So, the given series $$\sum_{k=1}^{\infty} \frac{k^{2}+k-1}{k^{4}+4 k^{2}-3}$$ converges by the Limit Comparison Test.

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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 useful tool in determining the convergence or divergence of an infinite series. It is especially handy when dealing with complicated fractions by comparing them to a simpler, known series. Here鈥檚 how you generally use it:
  • Select a comparison series that has known convergence properties, like a p-series.
  • Compute the limit of the ratio of the terms of the given series and the comparison series as the index approaches infinity.
  • If the limit is a positive finite number, both series either converge or diverge together.
In our original exercise, the Limit Comparison Test was applied to simplify the given series by forming a ratio with the p-series \( rac{1}{k^2}\). After simplifying and finding the limit to be 0, which is a valid finite number for this test, it was concluded that the original series converges similar to the comparison series.
p-series
A p-series is an important concept in calculus, which is critical when examining series convergence. A p-series is of the form: \[\sum_{k=1}^{\infty} \frac{1}{k^p}\] where \(p\) is a constant. The rules for convergence of a p-series are straightforward:
  • If \(p > 1\), the series converges.
  • If \(p \leq 1\), the series diverges.
p-series play a fundamental role in comparison tests because they serve as a basis to determine the behavior of more complex series. In the example provided, the series \(\sum_{k=1}^{\infty} \frac{1}{k^2}\) is a convergent p-series since \(p = 2\). This known behavior is what allows us to assert the convergence of more complex series when using comparison methods.
Comparison Test
The Comparison Test is another essential method for analyzing whether a series converges or diverges. Here, we closely relate our series with another series which has a known convergence behavior. The test comes in two variations:
  • Direct Comparison Test: Compare each term of the series directly with the terms of a known series. If each term of our series is smaller than a term of a known convergent series, then the series converges. Conversely, if each term is larger than a term of a divergent series, it diverges.
  • Limit Comparison Test: This is a variant used when direct comparison isn鈥檛 convenient. If the ratio of terms from our series and the known series approaches a finite number greater than zero, they both share the same convergence behavior.
In the original exercise, the Limit Comparison Test was preferred over the direct method, simplifying the process of determining convergence with the given fractions. The clearly defined steps made it easier to see the relationship between the more complex series and a simpler p-series.

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Most popular questions from this chapter

The Greek philosopher Zeno of Elea (who lived about 450 B.c.) invented many paradoxes, the most famous of which tells of a race between the swift warrior Achilles and a tortoise. Zeno argued The slower when running will never be overtaken by the quicker: for that which is pursuing must first reach the point from which that which is fleeing started, so that the slower must necessarily always be some distance ahead. In other words, by giving the tortoise a head start, Achilles will never overtake the tortoise because every time Achilles reaches the point where the tortoise was, the tortoise has moved ahead. Resolve this paradox by assuming that Achilles gives the tortoise a 1 -mi head start and runs \(5 \mathrm{mi} / \mathrm{hr}\) to the tortoise's \(1 \mathrm{mi} / \mathrm{hr}\). How far does Achilles run before he overtakes the tortoise, and how long does it take?

A tank is filled with 100 L of a \(40 \%\) alcohol solution (by volume). You repeatedly perform the following operation: Remove 2 L of the solution from the tank and replace them with 2 L of \(10 \%\) alcohol solution. a. Let \(C_{n}\) be the concentration of the solution in the tank after the \(n\) th replacement, where \(C_{0}=40 \% .\) Write the first five terms of the sequence \(\left\\{C_{n}\right\\}\). b. After how many replacements does the alcohol concentration reach \(15 \% ?\). c. Determine the limiting (steady-state) concentration of the solution that is approached after many replacements.

The Fibonacci sequence \(\\{1,1,2,3,5,8,13, \ldots\\}\) is generated by the recurrence relation \(f_{n+1}=f_{n}+f_{n-1},\) for \(n=1,2,3, \ldots,\) where \(f_{0}=1, f_{1}=1\). a. It can be shown that the sequence of ratios of successive terms of the sequence \(\left\\{\frac{f_{n+1}}{f_{n}}\right\\}\) has a limit \(\varphi .\) Divide both sides of the recurrence relation by \(f_{n},\) take the limit as \(n \rightarrow \infty,\) and show that \(\varphi=\lim _{n \rightarrow \infty} \frac{f_{n+1}}{f_{n}}=\frac{1+\sqrt{5}}{2} \approx 1.618\). b. Show that \(\lim _{n \rightarrow \infty} \frac{f_{n-1}}{f_{n+1}}=1-\frac{1}{\varphi} \approx 0.382\). c. Now consider the harmonic series and group terms as follows: $$\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\left(\frac{1}{4}+\frac{1}{5}\right)+\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)$$ $$+\left(\frac{1}{9}+\cdots+\frac{1}{13}\right)+\cdots$$ With the Fibonacci sequence in mind, show that $$\sum_{k=1}^{\infty} \frac{1}{k} \geq 1+\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{3}{8}+\frac{5}{13}+\cdots=1+\sum_{k=1}^{\infty} \frac{f_{k-1}}{f_{k+1}}.$$ d. Use part (b) to conclude that the harmonic series diverges. (Source: The College Mathematics Journal, 43, May 2012)

Use Exercise 89 to determine how many terms of each series are needed so that the partial sum is within \(10^{-6}\) of the value of the series (that is, to ensure \(\left|R_{n}\right|<10^{-6}\) ). Functions defined as series Suppose a function \(f\) is defined by the geometric series \(f(x)=\sum_{k=0}^{\infty}(-1)^{k} x^{k}\) a. Evaluate \(f(0), f(0.2), f(0.5), f(1),\) and \(f(1.5),\) if possible. b. What is the domain of \(f ?\)

A fallacy Explain the fallacy in the following argument. Let \(x=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots\) and \(y=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots \cdot\) It follows that \(2 y=x+y\) which implies that \(x=y .\) On the other hand, $$ x-y=\underbrace{\left(1-\frac{1}{2}\right)}_{>0}+\underbrace{\left(\frac{1}{3}-\frac{1}{4}\right)}_{>0}+\underbrace{\left(\frac{1}{5}-\frac{1}{6}\right)}_{>0}+\cdots>0 $$ is a sum of positive terms, so \(x>y .\) Therefore, we have shown that \(x=y\) and \(x>y\)
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