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Chapter 9: Problem 41

Using the Root Test In Exercises \(39-52,\) use the Root Test to determine the convergence or divergence of the series. $$\sum_{n=1}^{\infty}\left(\frac{3 n+2}{n+3}\right)^{n}$$

Short Answer

Expert verified
The series \(\sum_{n=1}^{\infty}\left(\frac{3n+2}{n+3}\right)^{n}\) diverges.

Step by step solution

01

Understanding the Root Test

The Root Test is a method used to test for the convergence/divergence of an infinite series. According to the Root Test, if we have \(\lim_{n \to \infty} \sqrt[n]{|a_n|} = L\), then the series \(\sum_{n}a_n\) will converge if \(L<1\), diverge if \(L>1\) and be inconclusive if \(L=1\)
02

Apply the Root Test

Let's apply the Root Test to the given series. We need to calculate the limit \(\lim_{n \to \infty} \sqrt[n]{\left|\frac{3n+2}{n+3}\right|^{n}}\). This simplifies to \(\lim_{n \to \infty} \frac{3n+2}{n+3}\).
03

Calculate the Limit

To find the limit \(\lim_{n \to \infty} \frac{3n+2}{n+3}\), we can divide the numerator and the denominator by n, the highest power in the denominator, to get \(\lim_{n \to \infty} \frac{3 + \frac{2}{n}}{1 + \frac{3}{n}}\). As \(n\) approaches infinity, \(\frac{2}{n}\) and \(\frac{3}{n}\) approach 0, so the limit is \(L = \frac{3 + 0}{1 + 0} = 3\).
04

Determine Convergence/Divergence

Since the limit \(L = 3\) is greater than 1, by the Root Test, the series \(\sum_{n=1}^{\infty}\left(\frac{3n+2}{n+3}\right)^{n}\) diverges.

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Key Concepts

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

Convergence
The concept of convergence in mathematics, particularly with series, describes a situation where the sum of infinite terms approaches a definite value as more terms are added. When dealing with infinite series, checking for convergence is crucial.
  • A series is said to converge if the sum of its terms moves towards a specific, finite number.
  • Common tests for checking convergence include the Root Test, Ratio Test, and Integral Test.
  • Convergence is essential because it indicates whether the series accumulates to a meaningful result or not. If it converges, it means you can find a finite sum.
Understanding convergence helps determine if a series represents a stable and usable value, especially in mathematical analysis and applied sciences. It ensures computations relying on the series remain consistent and reliable over time.
Divergence
Divergence is the opposite of convergence. In the realm of infinite series, this term is used when the sums of the series do not settle towards a fixed, finite number as more terms are added. Instead, they continue to grow without bound or oscillate without settling.
  • A diverging series will not settle on a finite sum, rendering it useless for calculations intending to achieve a stable value.
  • The Root Test, among others, is a tool to identify divergence by checking the limit of the sequence鈥檚 terms.
  • If the limit result during the Root Test is greater than one, as in the case of our example, the series is said to diverge.
Recognizing divergence is crucial when analyzing mathematical series, as it highlights scenarios where further investigations or alterations are required to find usable mathematical results.
Infinite Series
An infinite series is a sum of infinitely many terms, expressed as \(\sum_{n=1}^{fty} a_n\). In the world of mathematics, such series often model real-world phenomena and complex mathematical processes.
  • Infinite series can be daunting but are incredibly useful in representing complex ideas concisely and computationally.
  • Understanding whether an infinite series converges or diverges is key to determining its usefulness in calculations and predictions.
  • Typical evaluations involve determining the characteristic behavior of the series terms as they extend to infinity.
By exploring infinite series, mathematicians and scientists can unravel the behavior of dynamic systems and effectively describe phenomena in mathematics, physics, and engineering.
Limit Calculation
Calculating limits is a fundamental process in mathematics, particularly when dealing with series and sequences. It helps determine the behavior of terms as they approach infinity.
  • Limits help identify the eventual behavior of a function or series, providing essential information for making conclusions about convergence or divergence.
  • In the Root Test example, we calculated the limit to be 3 by simplifying the expression \(\lim_{n \to \infty} \frac{3n+2}{n+3}\) by factoring and canceling relevant terms.
  • This calculation showed how the series behaves as we reach larger values of \(n\), determining the final decision regarding divergence.
Grasping the concept of limits is critical, especially in calculus, as it forms the foundation for understanding continuity, integrals, derivatives, and the overall behavior of functions.

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Most popular questions from this chapter

Differential Equation In Exercises \(59-64,\) show that the function represented by the power series is a solution of the differential equation. $$y=\sum_{n=0}^{\infty} \frac{x^{2 n}}{2^{n} n !}, \quad y^{\prime \prime}-x y^{\prime}-y=0$$

Finding the Interval of Convergence In Exercises \(41-44,\) find the interval of convergence of the power series, where \(c>0\) and \(k\) is a positive integer. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-c)^{n}}{n c^{n}}$$

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 \(x=0 . )\) $$\int_{0}^{1} e^{-x^{3}} d x$$

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.) $$\sum_{n=0}^{\infty} \frac{(-1)^{n} n !(x-5)^{n}}{3^{n}}$$

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.) $$\sum_{n=1}^{\infty}\left[\frac{2 \cdot 4 \cdot 6 \cdot \cdot \cdot 2 n}{3 \cdot 5 \cdot 7 \cdot \cdot \cdot(2 n+1)}\right] x^{2 n+1}$$
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