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Chapter 8: Problem 19

State what conclusion, if any, may be drawn from the Divergence Test. $$ \sum_{n=1}^{\infty} \sin (1 / n) $$

Short Answer

Expert verified
The Divergence Test is inconclusive for \( \sum_{n=1}^{\infty} \sin(1/n) \).

Step by step solution

01

Recognize the Divergence Test

The Divergence Test, also known as the nth-term test for divergence, states: If \( \lim_{{n \to \infty}} a_n eq 0 \), or the limit does not exist, then the series \( \sum_{n=1}^{\infty} a_n \) diverges. However, if \( \lim_{{n \to \infty}} a_n = 0 \), the test is inconclusive.
02

Identify the Sequence Term

Identify the term of the sequence you are testing for divergence: \( a_n = \sin(1/n) \).
03

Evaluate the Limit of the Sequence Term

Calculate \( \lim_{{n \to \infty}} \sin(1/n) \). As \( n \to \infty \), \( 1/n \to 0 \). Since \( \sin(x) \to x \) as \( x \to 0 \), it follows that \( \sin(1/n) \to 0 \). Thus, \( \lim_{{n \to \infty}} \sin(1/n) = 0 \).
04

Conclusion Using the Divergence Test

Based on the result from Step 3, since \( \lim_{{n \to \infty}} \sin(1/n) = 0 \), the Divergence Test is inconclusive. It does not provide any information about the convergence or divergence of the series \( \sum_{n=1}^{\infty} \sin(1/n) \).

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Key Concepts

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

Series Convergence
In mathematics, especially in calculus, understanding whether a series converges or diverges is vital. A series is essentially a sum of terms in a sequence. When we ask if a series converges, we're questioning whether the sum of its infinite terms arrives at a finite number. If it does, the series is convergent; if not, it is divergent.

The Divergence Test, or the nth-term test, is one of the simplest methods to check for divergence. It states that if the limit of the sequence terms is not zero, the series must diverge. However, this test only helps to identify divergence and can鈥檛 confirm convergence. If the limit is zero, as in the case of our exercise, where \( \lim_{{n \to \infty}} \sin(1/n) = 0 \), it simply means the Divergence Test is inconclusive. So, additional tests or methods need to be employed to check for convergence.
Limit Evaluation
Evaluating limits is a key skill when working with sequences and series. Limits help us understand the behavior of a sequence as it approaches infinity. When evaluating \( \lim_{{n \to \infty}} \sin(1/n) \), it's constructive to note the behavior of \( \sin(x) \) as \( x \to 0 \).

Because \( \sin(x) \to x \) when \( x \to 0 \), and \( 1/n \to 0 \) as \( n \to \infty \), it follows that \( \sin(1/n) \to 0 \). This evaluation confirms that the limit of the sequence term \( a_n = \sin(1/n) \) is zero. This limit informs us that the Divergence Test is not helpful in this scenario. To fully determine whether a series converges, further analysis beyond the Divergence Test, such as using the Comparison Test or the Integral Test, might be essential.
Sequence Term Identification
Identifying the core term in a sequence is like discovering the building block of a series. For our particular exercise, the term \( a_n = \sin(1/n) \) must be identified and analyzed. Knowing this term allows us to evaluate its behavior over large values of \( n \).

The sequence of interest, \( \{\sin(1/n)\} \), involves applying the formula of the sine function to a reciprocal term. Understanding how the term changes informs us about the potential convergence or divergence of the series. This term must be processed through relevant tests to reach conclusions about the series. Observing that \( \sin(1/n) \to 0 \) gives us limited information initially but is crucial for initial assessments in calculus.

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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. The secant function has a known power series expansion that begins $$ \sec (x)=1+\frac{1}{2} x^{2}+\frac{5}{24} x^{4}+\frac{61}{720} x^{6} \ldots $$ The sine function has a known power series expansion that begins $$ \sin (x)=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !} \ldots $$ The tangent function has a known power series expansion that begins $$ \tan (x)=x+\frac{1}{3} x^{3}+\frac{2}{15} x^{5}+\frac{17}{315} x^{7}+\cdots $$ Verify the Cauchy product formula for \(\tan (x)=\sin (x)\) \(\sec (x)\) up to the \(x^{7}\) term.

Use the Uniqueness Theorem to determine the coefficients \(\left\\{a_{n}\right\\}\) of the solution \(y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the given initial value problem. \(d y / d x=2 y \quad y(0)=3\)

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. \(\sum_{n=1}^{\infty}(-1)^{n} \ln (2+1 / n)\)

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 \(S_{7}(x)\).

Consider the initial value problem $$ \frac{d y}{d x}=2-x-y, \quad y(0)=1 $$ \(\begin{array}{llll}\text { a. Calculate the power series } & \text { expansion }\end{array}\) \(y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) 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 \([-2,2] \times\) [-10,1.7] c. The exact solution to the initial value problem is \(y(x)=3-x-2 e^{-x},\) 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. To see the improvement in accuracy that results from using more terms in a partial sum, replace the graph of \(S_{3}(x)\) with that of \(S_{7}(x)\)
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