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Chapter 9: Problem 40

Find the radius of convergence and the interval of convergence. $$\sum_{k=2}^{\infty}(-1)^{k+1} \frac{x^{k}}{k(\ln k)^{2}}$$

Short Answer

Expert verified
The radius of convergence is 1 and the interval of convergence is [-1, 1].

Step by step solution

01

Analyze the General Term

Identify the general term of the given series. The general term is \((-1)^{k+1} \frac{x^{k}}{k(\ln k)^{2}}\). Here, \(a_k = \frac{(-1)^{k+1} x^{k}}{k(\ln k)^{2}}\), where \(k \geq 2\).
02

Apply the Ratio Test

To determine the radius of convergence, we will apply the ratio test. Compute the ratio of consecutive terms,\[L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| = \lim_{k \to \infty} \left| \frac{(-1)^{k+2} x^{k+1} / ((k+1)(\ln(k+1))^2)}{(-1)^{k+1} x^{k} / (k(\ln k)^2)} \right|\]Simplify to get:\[L = \lim_{k \to \infty} \left| x \cdot \frac{k(\ln k)^2}{(k+1)(\ln(k+1))^2} \right| = |x| \cdot \lim_{k \to \infty} \frac{k(\ln k)^2}{k(\ln k)^2 + O(k)}\]As \(k \to \infty\), this limit simplifies to \(|x| \cdot 1 = |x|\). The series converges when \(|x| < 1\).
03

Determine the Radius of Convergence

From the ratio test result, we find that the series converges absolutely when \(|x| < 1\). Therefore, the radius of convergence \(R\) is \(1\).
04

Check the Interval of Convergence

To find the interval of convergence, test the endpoints \(x = -1\) and \(x = 1\).- For \(x = -1\): The series becomes \(\sum_{k=2}^\infty \frac{(-1)^{k+1}(-1)^k}{k(\ln k)^2} = \sum_{k=2}^\infty \frac{1}{k(\ln k)^2}\), which is a convergent series by the integral test.- For \(x = 1\): The series becomes \(\sum_{k=2}^\infty \frac{(-1)^{k+1}}{k(\ln k)^2}\), which converges by the Alternating Series Test because the terms decrease in absolute value and tend to zero.Therefore, the series converges for \(-1 \leq x \leq 1\).
05

Combine Results

The radius of convergence is \(R = 1\), and the interval of convergence is \([-1, 1]\).

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Key Concepts

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

Interval of Convergence
The interval of convergence is essential in determining where a series converges. It showcases the values of \(x\) for which the power series converges.- To find this interval, first calculate the radius of convergence (often denoted as \(R\)).- The interval is typically of the form \((-R, R)\) but may include endpoints upon further testing.For the given series \(\sum_{k=2}^{\infty} (-1)^{k+1} \frac{x^{k}}{k(\ln k)^{2}}\),- After applying the ratio test, we found that \(|x| < 1\) for convergence.- Testing the endpoints by substituting \(x = -1\) and \(x = 1\) reveals that both endpoints allow for convergence. Consequently, the interval of convergence is \([-1, 1]\). Explore each endpoint using the necessary tests for more confidence in your conclusion.
Ratio Test
The ratio test is a method used to determine the convergence of a series.- The test analyzes the limit of the absolute value of the ratio of consecutive terms \(|a_{k+1}/a_k|\).- If the limit \(L < 1\), the series converges absolutely.- If \(L > 1\) or \(L = \infty\), the series diverges.- If \(L = 1\), the test is inconclusive.For our series, the ratio test shows \(L = |x|\). Therefore, the convergence condition \(|x| < 1\) is satisfied, giving us the radius of convergence as \(R = 1\). This simplifies the interval determination and offers a crucial insight into where the series converges firmly.
Alternating Series Test
The alternating series test helps us judge the convergence of series where terms alternate in sign.An alternating series \(\sum (-1)^{k} b_k\) converges if:
  • The absolute value of the terms \(b_k\) decreases monotonically.
  • The limit of the terms \(b_k\) as \(k\) approaches infinity is zero.
In our series for \(x = 1\), it simplifies to \(\sum (-1)^{k+1} \frac{1}{k(\ln k)^2}\). It passes both conditions of diminishing terms and a limit of zero. Thus, we ensure convergence at this endpoint based on the alternating series test, making it a useful tool for understanding behavior at specific \(x\) values.
Integral Test
The integral test is applicable to series with positive, continuous, and decreasing terms. It links the convergence of a series to the corresponding improper integral.Given a series \(\sum a_k\), the integral test says:- If \(\int_{a}^{\infty} f(x)\,dx\) converges, then so does \(\sum a_k\).- If \(\int_{a}^{\infty} f(x)\,dx\) diverges, \(\sum a_k\) does too.For \(x = -1\), the series becomes \(\sum \frac{1}{k(\ln k)^2}\). Using the integral test, we examine:\[\int_{2}^{\infty} \frac{1}{x(\ln x)^2}\, dx\]. This integral converges, thus showing the series converges for \(x = -1\), proving the series converges at this endpoint. The integral test is highly valuable in conjunction with other tests to understand full interval behavior.

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Most popular questions from this chapter

Use the Remainder Estimation Theorem to find an interval containing \(x=0\) over which \(f(x)\) can be approximated by \(p(x)\) to three decimal-place accuracy throughout the interval. Check your answer by graphing \(|f(x)-p(x)|\) over the interval you obtained. $$f(x)=\sin x ; p(x)=x-\frac{x^{3}}{3 !}$$

The limit of an indeterminate form as \(x \rightarrow x_{0}\) can sometimes be found by expanding the functions involved in Taylor series about \(x=x_{0}\) and taking the limit of the series term by term. Use this method to find the limits in these exercises. (a) \(\lim _{x \rightarrow 0} \frac{1-\cos x}{\sin x}\) (b) \(\lim _{x \rightarrow 0} \frac{\ln \sqrt{1+x}-\sin 2 x}{x}\)

Let \(\sum a_{k}\) and \(\sum b_{k}\) be series with positive terms. Prove: (a) If \(\lim _{k \rightarrow+\infty}\left(a_{k} / b_{k}\right)=0\) and \(\sum b_{k}\) converges, then \(\sum a_{k}\) converges. (b) If \(\lim _{k \rightarrow+\infty}\left(a_{k} / b_{k}\right)=+\infty\) and \(\sum b_{k}\) diverges, then \(\sum a_{k}\) diverges.

Show that for all real values of \(x\) \(\sin x-\frac{1}{2} \sin ^{2} x+\frac{1}{4} \sin ^{3} x-\frac{1}{8} \sin ^{4} x+\cdots=\frac{2 \sin x}{2+\sin x}\)

The limit of an indeterminate form as \(x \rightarrow x_{0}\) can sometimes be found by expanding the functions involved in Taylor series about \(x=x_{0}\) and taking the limit of the series term by term. Use this method to find the limits in these exercises. (a) \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) (b) \(\lim _{x \rightarrow 0} \frac{\tan ^{-1} x-x}{x^{3}}\)
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