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Chapter 10: Problem 35

In Exercises \(31-38,\) use the \(n\) th-Term Test for divergence to show that the series is divergent, or state that the test is inconclusive. $$\sum_{n=1}^{\infty} \cos \frac{1}{n}$$

Short Answer

Expert verified
The series diverges by the nth-Term Test, as the limit is 1.

Step by step solution

01

State the nth-Term Test for Divergence

The nth-term test for divergence states that if \( \lim_{{n \to \infty}} a_n eq 0 \), then the series \( \sum_{n=1}^{\infty} a_n \) is divergent. If \( \lim_{{n \to \infty}} a_n = 0 \), the test is inconclusive.
02

Identify the terms of the series

The terms of the series are given by \( a_n = \cos \frac{1}{n} \).
03

Evaluate the limit of the sequence terms

Compute \( \lim_{{n \to \infty}} \cos \frac{1}{n} \). Since \( \frac{1}{n} \to 0 \) as \( n \to \infty \), and \( \cos x \to 1 \) as \( x \to 0 \), it follows that \( \lim_{{n \to \infty}} \cos \frac{1}{n} = 1 \).
04

Apply the nth-Term Test

Since \( \lim_{{n \to \infty}} \cos \frac{1}{n} = 1 eq 0 \), we can conclude that the nth-term test shows the series \( \sum_{n=1}^{\infty} \cos \frac{1}{n} \) is divergent.

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Key Concepts

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

Divergent Series
Understanding divergent series is a key part of calculus and analysis. A series can be thought of as an infinite sum of terms. When the sum of all these terms doesn't tend to a finite limit, the series is called divergent.

This occurs when a series grows infinitely large or oscillates without approaching any fixed value.

Here's what you should know about divergent series:
  • A divergent series cannot be assigned a numerical value in the usual sense.
  • Even though individual terms might get smaller, the series sum keeps getting larger or oscillates.
  • The nth-Term Test for Divergence is a way to identify some divergent series.
For a series to diverge, using the nth-term test, its term's limit as it goes to infinity cannot be zero. If the limit is anything other than zero, then the series is divergent.
Limit of a Sequence
A fundamental concept when dealing with series and sequences is the limit of a sequence. The limit helps determine the behavior of a sequence as the number of terms grows larger and larger.

The limit of a sequence \( a_n \), written as \( \lim_{{n \to \infty}} a_n \), describes what value \( a_n \) gets closer and closer to, as \( n \) becomes infinitely large.

Here鈥檚 how you can understand limits further:
  • If \( \lim_{{n \to \infty}} a_n = L \), then eventually terms in the sequence get arbitrarily close to \( L \).
  • If \( \lim_{{n \to \infty}} a_n eq 0 \), many tests like the nth-Term Test conclude the series diverges.
  • Limits help us decide if series converge (approach a sum) or diverge.
In our example, we found \( \lim_{{n \to \infty}} \cos \frac{1}{n} = 1 \), indicating divergence, as the term's limit isn't zero.
Trigonometric Functions
Trigonometric functions, like cosine, often appear in series problems. Understanding their behaviors is vital in solving related mathematical problems.

For the cosine function, specifically:
  • It oscillates between -1 and 1.
  • As \( x \to 0 \), \( \cos x \to 1 \).
  • This property helps determine the limits involving the cosine function.
In the context of the series \( \sum_{n=1}^{\infty} \cos \frac{1}{n} \):

The term \( \cos \frac{1}{n} \) approaches 1 as \( n \to \infty \). Thus, each term of the series doesn't approach 0, crucial for determining divergence through the nth-Term Test.

Understanding trigonometric function properties can vastly improve your comprehension of how they interact in sequences and series.

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Most popular questions from this chapter

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}.$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation \begin{equation}e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x).\end{equation} Differentiate the right-hand side of this equation to show that \begin{equation}\frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x}.\end{equation} Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.

The sequence \(\\{n /(n+1)\\}\) has a least upper bound of 1 Show that if \(M\) is a number less than \(1,\) then the terms of 1 \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\} .\)

Uniqueness of limits Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\)

Determine if the sequence is monotonic and if it is bounded. $$ a_{n}=\frac{2^{n} 3^{n}}{n !} $$
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