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Chapter 9: Problem 23

Using the Limit Comparison Test In Exercises \(17-26,\) use the Limit Comparison Test to determine the convergence or divergence of the series. $$\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n^{2}+1}}$$

Short Answer

Expert verified
The given series \(\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n^{2}+1}}\) converges.

Step by step solution

01

Choose a Series for Comparison

Choose a series for comparison. Choose the series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) for comparison. This series is known to be convergent.
02

Limit Comparison Test formula

The limit comparison test states that if the limit as n approaches infinity of the ratio of the n-th term of two series is a finite value greater than zero, both series will either converge or diverge. Let's set up the limit formula: \(\displaystyle\lim_{{n} \to \infty} \frac{a_n}{b_n} = c,\) where \(a_n) is the n-th term of the original series and \(b_n\) is the term of the comparison series.
03

Evaluate the Limit

Evaluate the limit which is given by \(\displaystyle\lim_{{n} \to \infty} \frac{n^{2}}{n \sqrt{n^{2}+1}}=\displaystyle\lim_{{n} \to \infty} \frac{n^{2}}{n^{2}}=\displaystyle\lim_{{n} \to \infty} \frac{1}{\sqrt{1+\frac{1}{n^{2}}}}=1.\)
04

Draw Conclusion

Since the limit is finite and greater than zero, and we know that the series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) converges, then by the Limit Comparison Test, the given series \(\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n^{2}+1}}\) also converges.

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Key Concepts

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

Convergence
When we talk about convergence in the context of a series, we mean that the sum of its terms approaches a specific number as you add more and more terms. A series converges if there is a finite limit to the sum of its terms. This concept is crucial in determining the behavior of infinite series.
  • A convergent series will not "blow up" or approach infinity as we add terms.
  • The series in the step-by-step solution, \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), is a classic example of a convergent series.
  • If a series converges, it means calculating the sum gives us a meaningful and finite result, no matter how many terms we calculate.
Understanding convergence helps us predict whether calculations involving series will yield useful results without requiring infinite computation.
Divergence
Divergence is the opposite of convergence. When a series diverges, the sum of its terms does not settle on a finite number as we add more terms. It might grow without bound or oscillate endlessly without settling down.
  • Divergent series are tricky because they do not provide a finite sum.
  • The famous series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is an example of a divergent series, known as the harmonic series.
  • Determining divergence helps us avoid using series that do not sum to a specific number, guiding us to reason about their behavior.
Recognizing divergence is essential for understanding the limitations of mathematical models involving series.
Infinite Series
An infinite series is simply a sum composed of an endless number of terms. These series are expressed with the summation notation \(\sum\), where the terms follow a pattern based on the sequence \(a_n\).
  • Infinite series can converge or diverge, as discussed in earlier sections.
  • Not all infinite series behave the same way; some grow, shrink, or oscillate.
  • The challenge is determining what happens as we take the limit as \(n\) approaches infinity.
The concept of infinite series is foundational in calculus and analysis, allowing us to work with functions, models, and phenomena that extend indefinitely.
Comparison Test
The Comparison Test is a handy method in mathematics to determine whether a series converges or diverges by comparing it to another series with known behavior. There are two primary versions: the Direct Comparison and the Limit Comparison Test.
  • In the Direct Comparison Test, we compare the terms of the series with another known series directly.
  • The Limit Comparison Test involves computing the limit of the ratio of the terms of the two series.
  • If the limit of the ratio exists and is a positive finite number, both series share the same convergence behavior.
The Limit Comparison Test in the original exercise helped us determine that \(\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n^{2}+1}}\) converges, by comparing it to the convergent series \(\sum_{n=1}^{\infty} \frac{1}{n^2}\). This method is powerful because it simplifies complex problems by linking them to already understood series.

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Most popular questions from this chapter

Approximating an Integral In Exercises \(63-70\) , use a power series to approximate the value of the definite integral with an error of less than \(0.0001 .\) (In Exercises 65 and \(67,\) assume that the integrand is defined as 1 when \(x=0 . )\) $$\int_{0.1}^{0.3} \sqrt{1+x^{3}} d x$$

Finding the Interval of Convergence In Exercises \(15-38\) , find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{n+1}$$

Finding Intervals of Convergence In Exercises \(49-52\) , find the intervals of convergence of $$f(x),\left(\text { b) } f^{\prime}(x),\left(\text { c) } f^{\prime \prime}(x), \text { and }\left(\text { d) } \int f(x) d x\right.\right.\right.$$ (Be sure to include a check for convergence at the endpoints of the intervals.) $$f(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{n+1}$$

$$\begin{array}{l}{\text { Investigation The interval of convergence of the geometric }} \\ {\text { series } \sum_{n=0}^{\infty}\left(\frac{x}{4}\right)^{n} \text { is }(-4,4)}\\\ {\text { (a) Find the sum of the series when } x=\frac{3}{2} . \text { Use a graphing }} \\ {\text { utility to graph the first six terms of the sequence of partial }} \\ {\text { sums and the horizontal line representing the sum of the }} \\ {\text { series. }} \\ {\text { (b) Repeat part (a) for } x=-\frac{5}{2} \text { . }}\\\ {\text { (c) Write a short paragraph comparing the rates of convergence }} \\ {\text { of the partial sums with the sums of the series in parts (a) }} \\ {\text { and (b). How do the plots of the partial sums differ as they }} \\ {\text { converge toward the sum of the series? }} \\ {\text { (d) Given any positive real number } M, \text { there exists a positive }} \\\ {\text { integer } N \text { such that the partial sum }}\end{array}$$ $$\sum_{n=0}^{N}\left(\frac{5}{4}\right)^{n}>M$$

$$\begin{array}{l}{\text { Proof Prove that the power series }} \\\ {\sum_{n=0}^{\infty} \frac{(n+p) !}{n !(n+q) !} x^{n}} \\ {\text { has a radius of convergence of } R=\infty \text { when } p \text { and } q \text { are }} \\ {\text { positive integers. }}\end{array}$$
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