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Chapter 10: Problem 33

Which of the series converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$\sum_{n=1}^{\infty} n \sin \frac{1}{n}$$

Short Answer

Expert verified
The series diverges because \( \lim_{n \to \infty} n \sin \frac{1}{n} \approx 1 \neq 0 \).

Step by step solution

01

Identify the Form of the Series

The series given is \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \). Recognize that as \( n \) becomes large, \( \sin \frac{1}{n} \approx \frac{1}{n} \) since \( \sin x \approx x \) when \( x \) is near zero.
02

Simplify the Expression

Approximate \( n \sin \frac{1}{n} \approx n \cdot \frac{1}{n} = 1 \). Thus, the series closely resembles \( \sum_{n=1}^{\infty} 1 \).
03

Determine Convergence of the Resembling Series

The series \( \sum_{n=1}^{\infty} 1 \) is a harmonic series which diverges. This is because adding 1 infinitely many times will not tend to a finite sum.
04

Use the Divergence Test

By the divergence test, if \(\lim_{n \to \infty} a_n eq 0\), where \( a_n = n \sin \frac{1}{n} \approx 1 \), the series \( \sum a_n \) diverges. Since \( \lim_{n \to \infty} n \sin \frac{1}{n} = 1 eq 0 \), the original series diverges.

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Key Concepts

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

Understanding the Divergence Test
The Divergence Test, also known as the Test for Divergence, is a straightforward and often the first test used when checking series convergence. It acts as an initial filter to quickly identify divergent series. Here's how it works:

  • For a series \( \sum a_n \), examine the limit of the sequence \( \{a_n\} \) as \( n \) approaches infinity.
  • If \( \lim_{n \to \infty} a_n eq 0 \), the series diverges.
  • If \( \lim_{n \to \infty} a_n = 0 \), the test is inconclusive. The series may converge or diverge.
This test is particularly useful for series where the terms do not tend to zero. In the original exercise, we considered \( \lim_{n \to \infty} n \sin \frac{1}{n} \). Since this limit equals 1, which is not zero, the series \( \sum n \sin \frac{1}{n} \) diverges by the Divergence Test.
Exploring the Harmonic Series
The harmonic series is one of the most famous examples of a divergent series. It is given by \( \sum_{n=1}^{\infty} \frac{1}{n} \). This series diverges, meaning it grows indefinitely as more terms are added.

Here's a simple way to understand why it diverges:
  • Group the terms in such a way to see the growth clearly: \( \frac{1}{1} + \left(\frac{1}{2} + \frac{1}{3}\right) + \left(\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7}\right) + \ldots \)
  • Each group of terms \((\frac{1}{2^k+1} + \cdots + \frac{1}{2^{k+1}})\) sums to more than 0.5.
The harmonic series slowly yet surely accumulates without bound. In the original exercise, the series \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \) resembled a constant sum series due to simplification, hinting divergence akin to that of the harmonic series.
The Power of the Limit Comparison Test
The Limit Comparison Test is a powerful tool when dealing with series that resembles another series whose convergence characteristics we know. It effectively tells us if two series "behave" the same way at infinity.

Here鈥檚 how you apply it:
  • Suppose you have two positive term series \( \sum a_n \) and \( \sum b_n \).
  • If the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \), where \( 0 < c < \infty \), then both series either converge or diverge together.
In practical scenarios, one often uses this test by comparing a given series to a known benchmark series like the harmonic series. However, in the original exercise, since the series \( \sum n \sin \frac{1}{n} \) was simplified already to closely resemble a divergent harmonic form, the divergence conclusion was drawn straightforwardly using the Divergence Test.

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Most popular questions from this chapter

Does the series $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n^{2}}\right) $$ converge or diverge? Justify your answer.

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 \(L ?\) \(a_{1}=1, \quad a_{n+1}=a_{n}+\frac{1}{5^{n}}\)

If the terms of one sequence appear in another sequence in their given order, we call the first sequence a sub-sequence of the second. Prove that if two sub-sequences of a sequence \(\left\\{a_{n}\right\\}\) have different limits \(L_{1} \neq L_{2}\) then \(\left\\{a_{n}\right\\}\) diverges.

Assume that each sequence converges and find its limit. \(a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2}\)

Use the definition of convergence to prove the given limit. \(\lim _{n \rightarrow \infty} \frac{\sin n}{n}=0\)
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