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Chapter 11: Problem 18

In Exercises \(1-32,\) (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=0}^{\infty} \frac{n x^{n}}{4^{n}\left(n^{2}+1\right)} $$

Short Answer

Expert verified
Radius of convergence: \(R = 4\). Absolute convergence for \(-4 < x < 4\); conditional convergence possible at endpoints.

Step by step solution

01

Ratio Test for Radius of Convergence

To determine the radius of convergence, apply the Ratio Test. For the series \(\sum_{n=0}^{\infty} a_n \), where \(a_n = \frac{n x^n}{4^n (n^2 + 1)}\), the Ratio Test states that the series converges if \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\). Calculate:\[\frac{a_{n+1}}{a_n} = \frac{(n+1)x^{n+1}}{4^{n+1}((n+1)^2 + 1)} \cdot \frac{4^n(n^2 + 1)}{n x^n} = \frac{(n+1)x}{4(n^2+1)}\cdot \frac{n^2 + 1}{(n+1)^2 + 1}\].Simplifying, \(\frac{a_{n+1}}{a_n} \approx \frac{|x|}{4}\), as \(n \to \infty\). Thus, the series converges when \(\frac{|x|}{4} < 1\) or \(|x| < 4\). Therefore, the radius of convergence is \(R = 4\).
02

Interval of Convergence

With a radius of convergence \(R = 4\), determine the interval of convergence. Initially, we have \(-4 < x < 4\). We must check the endpoints \(x = -4\) and \(x = 4\) for convergence individually.
03

Test Endpoint \(x = -4\)

Substitute \(x = -4\) into the series:\[\sum_{n=0}^{\infty} \frac{n (-4)^n}{4^n(n^2+1)} = \sum_{n=0}^{\infty} \frac{n (-1)^n}{n^2+1}\].This series does not converge absolutely, as the terms \(\frac{n}{n^2+1}\) do not converge to zero. A more detailed test (such as the Alternating Series Test) would be needed to determine conditional convergence.
04

Test Endpoint \(x = 4\)

Substitute \(x = 4\) into the series:\[\sum_{n=0}^{\infty} \frac{n 4^n}{4^n(n^2+1)} = \sum_{n=0}^{\infty} \frac{n}{n^2+1}\].This series diverges by the Divergence Test since the terms \(\frac{n}{n^2+1}\) approach 0 slower than \(\frac{1}{n}\), whose series is known to diverge.
05

Absolute Convergence

The series converges absolutely for \(-4 < x < 4\). Within this interval, the original series is absolutely convergent because the corresponding terms \(\frac{n x^n}{4^n(n^2+1)}\) fulfill the criteria for absolute convergence according to the ratio test.
06

Conditional Convergence

From Step 3, the series at \(x = -4\) was found not to converge absolutely. Thus, conditional convergence needs more elaborate checks, such as applying the Alternating Series Test. If convergent under this test, the series would have conditional convergence at these endpoints, otherwise it does not.

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Key Concepts

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

Radius of Convergence
The radius of convergence is a fundamental concept when dealing with power series. For a power series given by the formula \(\sum_{n=0}^{\infty} a_n \), the radius of convergence is determined by applying the Ratio Test. The idea is to find the value of \(R\) within which the series converges, that is, all values of \(x\) will need to satisfy the condition \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\).
  • In the example, \(a_n = \frac{n x^n}{4^n (n^2 + 1)}\).
  • By calculating the limit of \(\left| \frac{a_{n+1}}{a_n} \right|\), we find \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \approx \frac{|x|}{4}\).
  • This means the series converges when \(|x| < 4\), giving a radius of convergence \(R = 4\).
This value represents the distance from the center of the series鈥攗sually at zero鈥攚here the series is guaranteed to converge absolutely.
Interval of Convergence
After identifying the radius of convergence \(R = 4\), it's crucial to determine the interval of convergence. This interval expresses the range of \(x\) values for which the series converges.
  • Initially, the interval appears as \(-4 < x < 4\), defined by the radius \(R = 4\).
  • However, we must test the convergence at the boundaries \(x = -4\) and \(x = 4\) separately.
Testing the endpoints is crucial, as they can either include or exclude the extremes of convergence based on how the series behaves at these points.
Ratio Test
The Ratio Test is a powerful tool used to determine the convergence of series. It works particularly well with power series.
  • For any series \(\sum a_n\), the Ratio Test focuses on the limit \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
  • If this limit is less than 1, the series converges absolutely.
  • If the limit exceeds 1, the series diverges.
  • When the limit precisely equals 1, the test is inconclusive, and further analysis is needed.
In our problem, we simplify to \(\left| \frac{a_{n+1}}{a_n} \right| \approx \frac{|x|}{4}\). This simplification allows us to determine the critical value for convergence, which here is \(|x| < 4\). Thus, illustrating the Ratio Test, it affirms the radius of convergence effectively.
Absolute Convergence
Absolute convergence deals with whether the series formed by taking the absolute values of the terms still converges.
  • When a series converges absolutely, it means that \(\sum |a_n|\) also converges.
  • This property is stronger than mere convergence, as absolute convergence implies convergence, but not vice versa.
For our series \(\sum_{n=0}^{\infty} \frac{n x^{n}}{4^{n}\left(n^{2}+1\right)}\), it converges absolutely whenever it converges by the Ratio Test in the interval \(-4 < x < 4\). Testing this interval reveals that within this range, all terms meet the standard for absolute convergence effectively. This knowledge aids in identifying regions where the series holds robustly.

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Most popular questions from this chapter

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{y \rightarrow 0} \frac{\tan ^{-1} y-\sin y}{y^{3} \cos y} $$

The value of \(\sum_{n=1}^{\infty} \tan ^{-1}\left(2 / n^{2}\right)\) a. Use the formula for the tangent of the difference of two angles to show that $$ \tan \left(\tan ^{-1}(n+1)-\tan ^{-1}(n-1)\right)=\frac{2}{n^{2}} $$ b. Show that $$ \sum_{n=1}^{N} \tan ^{-1} \frac{2}{n^{2}}=\tan ^{-1}(N+1)+\tan ^{-1} N-\frac{\pi}{4} $$ c. Find the value of \(\sum_{n=1}^{\infty} \tan ^{-1} \frac{2}{n^{2}}\)

Logistic difference equation The recursive relation $$ a_{n+1}=r a_{n}\left(1-a_{n}\right) $$ is called the logistic difference equation, and when the initial value \(a_{0}\) is given the equation defines the logistic sequence \(\left\\{a_{n}\right\\} .\) Throughout this exercise we choose \(a_{0}\) in the interval \(03.57\) . Choose \(r=3.65\) and calculate and plot the first 300 terms of \(\left\\{a_{n}\right\\} .\) Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of \(a_{n+1}\) from previous values of the sequence. g. For \(r=3.65\) choose two starting values of \(a_{0}\) that are close together, say, \(a_{0}=0.3\) and \(a_{0}=0.301 .\) Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for \(r=3.75 .\) Can you see how the plots look different depending on your choice of \(a_{0} ?\) We say that the logistic sequence is sensitive to the initial condition a_{0} .

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so does $$ \begin{array}{ll}{\text { a. } \sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. } \sum_{n=1}^{\infty} k a_{n} \quad(k \text { any number })}\end{array} $$

Uniqueness of least upper bounds Show that if \(M_{1}\) and \(M_{2}\) are least upper bounds for the sequence \(\left\\{a_{n}\right\\},\) then \(M_{1}=M_{2} .\) That is, a sequence cannot have two different least upper bounds.
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