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Which of the series, and which diverge? Use any method, and give reasons for your answers. $$\sum_{n=1}^{\infty} \sin \frac{1}{n}$$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Understand Series Convergence

The convergence of a series \(\sum_{n=1}^{\infty} a_n\) depends on whether the sequence of partial sums \(S_N = \sum_{n=1}^{N} a_n\) approaches a finite limit as \(N\) approaches infinity. If \(S_N\) approaches a finite limit, the series converges; otherwise, it diverges.
02

Apply the Limit Comparison Test

Given the series \(\sum_{n=1}^{\infty} \sin \frac{1}{n}\), we consider the behavior of \(\sin \frac{1}{n}\) for large \(n\). For small \(x\), \(\sin x\) is approximately \(x\). Therefore, when \(n\) is large, \(\sin \frac{1}{n} \approx \frac{1}{n}\). Compare this series with the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which is known to diverge.
03

Use Limit Test to Compare Terms

We compute the limit \(\lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}}\). Using the fact that \(\sin x \approx x\) as \(x\) approaches zero, this limit evaluates to 1: \[\lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = 1\]Since the limit is a finite non-zero number, the original series \(\sum_{n=1}^{\infty} \sin \frac{1}{n}\) behaves like the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which diverges.
04

Conclude on the Series

Since the comparison with the harmonic series indicates divergence and other convergence tests do not apply favorably here, we conclude that the series \(\sum_{n=1}^{\infty} \sin \frac{1}{n}\) diverges.

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Key Concepts

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

Divergence Tests
Divergence tests are crucial tools in determining whether a series converges or diverges. One simple yet effective method is the Test for Divergence. This test involves examining the terms of the series themselves. If a series \(\sum_{n=1}^{\infty} a_n\), with terms \(a_n\), does not satisfy \(\lim_{n \to \infty} a_n = 0\), then the series must diverge. Otherwise, if the limit equals zero, it suggests potential for convergence, but does not prove it.
Using the Test for Divergence, we first check the terms of our series. If \(a_n\) do not approach zero, divergence is certain. If they do approach zero, other tests are required, as convergence still needs to be confirmed.
Limit Comparison Test
The Limit Comparison Test helps us compare a given series with another series whose convergence we know well, such as the harmonic series. By examining the behavior of the series' terms, we can often draw conclusions about convergence or divergence of our original series.
Step through this with \(\sum_{n=1}^{\infty} \sin \frac{1}{n}\). Given that for small values, \(\sin x \approx x\), the terms can be approximated as \(\frac{1}{n}\) for large \(n\). Comparing this to the known divergent harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), we can utilize the Limit Comparison Test:
  • Calculate \(\lim_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}}\).
  • The limit equals 1, a finite and non-zero number.
This tells us that the behaviors of the two series are equivalent in terms of convergence properties, leading to the conclusion that our given series diverges.
Harmonic Series
The harmonic series is an essential reference point in the study of series convergence and divergence. It is given by:\(\sum_{n=1}^{\infty} \frac{1}{n}\).The harmonic series is well-known for its unique diverging behavior, despite its terms tending to zero as \(n\) increases.
Even though each individual term becomes smaller, their cumulative value does not sum to a finite number.
This serves as a classic example where terms approaching zero don't guarantee convergence, emphasizing a crucial understanding for series called the "harmonic divergence" concept.
Partial Sums
The concept of partial sums is fundamental to understanding convergence. For a series \(\sum_{n=1}^{\infty} a_n\), a partial sum \(S_N\) is the sum of the first \(N\) terms:\[S_N = \sum_{n=1}^{N} a_n\].Analyzing the sequence of partial sums provides insight into whether a series converges or diverges.
- If \(S_N\) approaches a finite value as \(N\) grows, the series is convergent.- Conversely, if \(S_N\) approaches infinity or fails to stabilize, divergence is indicated.
In practical terms, examining the growth or decline of \(S_N\) offers a tangible method to understand a series' behavior, tied directly to the convergence limits of its cumulative values.

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Most popular questions from this chapter

Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(\bar{x}=0\) Step 3: Calculate the ( \(n+1\) )st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=(\cos x)(\sin 2 x), \quad|x| \leq 2$$

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) \(a_{n}=\left(1+\frac{0.5}{n}\right)^{n}\)

Determine if the sequence is monotonic and if it is bounded. \(a_{n}=2-\frac{2}{n}-\frac{1}{2^{n}}\)

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

a. Show that the improper integral $$ \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant }) $$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

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