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Chapter 8: Problem 45

In each of Exercises \(45-60,\) determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{1}{\sqrt{n+10}}\)

Short Answer

Expert verified
The series converges conditionally.

Step by step solution

01

Identify the Type of Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{\sqrt{n+10}} \). It is an alternating series because of the \((-1)^n\) term. Our objective is to determine if it converges absolutely, converges conditionally, or diverges.
02

Test for Absolute Convergence

We first check for absolute convergence by considering the series with absolute values: \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+10}} \). This means we ignore the alternating part and test \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+10}} \).
03

Apply the Comparison Test for Absolute Convergence

Notice that the terms \( \frac{1}{\sqrt{n+10}} \sim \frac{1}{\sqrt{n}} \). Comparing with the p-series \( \sum \frac{1}{n^{1/2}} \), which diverges, since \(p = 1/2 \leq 1\), \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+10}} \) also diverges by the Comparison Test.
04

Test for Conditional Convergence Using Alternating Series Test

The Alternating Series Test states that if \( b_n = \frac{1}{\sqrt{n+10}} \) is positive, decreasing, and \( \lim_{n \to \infty} b_n = 0 \), then \( \sum_{n=1}^{\infty}(-1)^{n}b_n \) converges. Here, \( b_n > 0 \), \( b_n \) is decreasing, and \( \lim_{n \to \infty} \frac{1}{\sqrt{n+10}} = 0 \). Therefore, the alternating series converges conditionally.

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Key Concepts

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

Alternating Series
An alternating series is a type of series where the terms alternate in sign. This means the terms switch from positive to negative or vice versa as you progress through the series. For example, the series given as \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{\sqrt{n+10}} \] has an alternating component \[ (-1)^n \]. Such series are special and can be tested for convergence using the Alternating Series Test. The test requires:
  • The absolute value of the terms, \( b_n = \frac{1}{\sqrt{n+10}} \), must be decreasing.
  • \( \lim_{n \to \infty} b_n = 0 \).
If these conditions are met, the series converges. This alternating series converges conditionally since these conditions are satisfied.
Absolute Convergence
Absolute convergence is when the series of absolute values also converges. For a series \( \sum a_n \), absolute convergence means that \( \sum |a_n| \) converges. In the given series, testing the absolute convergence means evaluating \[ \sum_{n=1}^{\infty} \left| (-1)^{n} \frac{1}{\sqrt{n+10}} \right| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+10}}. \]This new series ignores the alternating sign, focusing on every term's positive value. In this case, the series does not converge absolutely because we compared it to the similar series \( \sum \frac{1}{\sqrt{n}} \), which diverges.
Conditional Convergence
Conditional convergence happens when a series converges, but does not converge absolutely. This applies to our series, \[ \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{\sqrt{n+10}}, \]which converges by the Alternating Series Test, while its absolute counterpart diverges.The process involves:
  • First checking the absolute convergence.
  • If that fails, applying the Alternating Series Test.
Since the absolute series fails to converge, but the alternating conditions hold, the series is conditionally convergent. This means the series itself converges, but the sum of the magnitudes of its terms would diverge.
Comparison Test
The Comparison Test is a useful tool for determining if a series converges or diverges by comparing it to a series that is already known to converge or diverge. For our series,\[ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+10}}, \]we used the Comparison Test to check its absolute convergence by comparing it with the p-series \[ \sum \frac{1}{n^{1/2}}, \]which is known to diverge, as \( p = \frac{1}{2} \leq 1 \). Since the terms of our series are similar in behavior to those of the divergent p-series (both grow without bound), our series also diverges. This comparison provided solid evidence that the series cannot converge absolutely.

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Most popular questions from this chapter

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. Suppose \(|x|<1 .\) Calculate the power series of \(h(x)=\) \(1 /(1-x)^{2}\) with base point 0 by using the method of Exercise 43\. Using \(f(x)=g(x)=1 /(1-x)=\sum_{n=0}^{\infty} x^{n},\) verify the Cauchy product formula for \(h=f \cdot g\) up to the \(x^{6}\) term.

Determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=2}^{\infty}(-1)^{n} \frac{1}{n \ln ^{3}(n)}\)

Use partial fractions to calculate the \(N^{\text {th }}\) partial sum \(S_{N}\) of the given series in closed form. Sum the series by finding \(\lim _{N \rightarrow \infty} S_{N}\). $$ \sum_{n=1}^{\infty} \frac{1}{(2 n+1)(2 n+3)} $$

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. Suppose that the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges on \((-R, R)\) to a function \(f(x)\) and that \(|f(x)| \geq k>0\) on that interval for some positive constant \(k\). Then, \(1 / f(x)\) also has a convergent power series expansion on \((-R, R) .\) Compute its coefficients in terms of the \(a_{n}\) 's. Hint: Set $$ \frac{1}{f(x)}=g(x)=\sum_{n=0}^{\infty} b_{n} x^{n} $$ Use the equation \(f(x) \cdot g(x)=1\) to solve for the \(b_{n}\) 's.

Consider the initial value problem $$ \frac{d y}{d x}=x^{2}+y, \quad y(0)=1 $$ \(\begin{array}{llll}\text { a. Calculate } & \text { the power } & \text { series } & \text { expansion }\end{array}\) \(y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the solution up to the \(x^{7}\) term. b. Using the coefficients you have calculated, plot \(S_{3}(x)=\sum_{n=0}^{3} a_{n} x^{n}\) in the viewing rectangle \([-3,3] \times\) [-11,44] c. The exact solution to the initial value problem is \(y(x)=3 e^{x}-x^{2}-2 x-2,\) as can be determined using the methods of Section 7.7 in Chapter 7 . Add the plot of the exact solution to the viewing window. From the two plots, we see that the approximation is fairly accurate for \(-1 \leq x \leq 1\), but the accuracy decreases outside this subinterval. d. When a partial sum \(S_{N}(x)\) is used to approximate an infinite series, an increase in the value of \(N\) requires more computation, but improved accuracy is the reward. To see the effect in this example, replace the graph of \(S_{3}(x)\) with that of \(S_{7}(x)\).
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