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Chapter 11: Problem 53

Suppose that \(a_{n} \rightarrow L .\) Show that if \(a_{n} \leq M\) for all \(n\), then \(L \leq M\)

Short Answer

Expert verified
To show that if \(a_n \rightarrow L\) and \(a_n \leq M\) for all n, then L 鈮� M, we assume for contradiction that L > M. Let \(\epsilon = L - M\), and use the definition of convergence to find a positive integer N such that for all \(n \geq N\), \(|a_n - L| < \epsilon\). This leads to the inequality \(a_n < M\) for all \(n \geq N\). However, this contradicts our assumption that L > M, so we conclude that L 鈮� M.

Step by step solution

01

Understand the convergence of a sequence

Recall that a sequence \(a_n\) converges to a limit L if for every \(\epsilon > 0\), there exists a positive integer N such that for all \(n \geq N\), we have \(|a_n - L| < \epsilon\). In other words, as n gets larger, the difference between the sequence terms and the limit L shrinks.
02

Assume L > M for contradiction

To prove that L 鈮� M, let's assume for the sake of contradiction that L > M. Then, we have \(L - M > 0\). Let \(\epsilon = L - M\). Since \(a_n\) converges to L, there exists a positive integer N such that for all \(n \geq N\), we have \(|a_n - L| < \epsilon\).
03

Check the inequality for n 鈮� N

We now consider the terms of the sequence for \(n \geq N\). By our assumption about the sequence, for such n, we have \(|a_n - L| < \epsilon\), which also means \(a_n - L < L - M\).
04

Simplify the inequality to find a contradiction

Rewrite the inequality: \(a_n - L < L - M \Rightarrow a_n < M\). But this is true for all \(n \geq N\), which contradicts our assumption that L > M.
05

Conclude that L 鈮� M

Since our assumption that L > M led to a contradiction, we conclude that L 鈮� M. Hence, if the sequence \(a_n\) converges to L and has an upper bound M for all terms, then the limit L is also less than or equal to M.

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Key Concepts

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

Limit of a Sequence
Understanding the limit of a sequence is crucial for analyzing how the sequence behaves as it progresses. From a mathematical point of view, a sequence \(a_n\) is said to have a limit \(L\) if the terms of the sequence get arbitrarily close to \(L\) as \(n\) becomes large. More formally, \(a_n\) converges to \(L\) if, for every \(\epsilon > 0\), there exists an \(N\) such that for all \(n \geq N\), the absolute difference \(\left| a_n - L \right|\) is less than \(\epsilon\). This definition ensures that after a certain point in the sequence, all terms are within an \(\epsilon\)-radius of \(L\), like an invisible bubble around \(L\).

For students to visually grasp the concept, it is often helpful to imagine the terms of the sequence on a number line getting closer and closer to the point representing the limit. As \(n\) increases, the numbers representing the sequence terms cluster more tightly around the limit, symbolizing convergence.
Proof by Contradiction
Proof by contradiction is a powerful logical tool used to establish the truth of a statement by first assuming that the opposite is true, and then showing that this leads to an illogical or impossible situation (a contradiction). When it comes to sequences, if you're trying to prove that a certain limit \(L\) must adhere to a particular condition鈥攕ay, being less than or equal to a given upper bound \(M\)鈥攜ou might start by assuming the opposite, that \(L > M\).

If this assumption leads directly to a situation that contradicts known information about the sequence or the properties of numbers, then the original assumption must be false. Therefore, by the principle of contradiction, the opposite of the assumption鈥攐ur original claim that \(L \leq M\)鈥攎ust be true. This method becomes an essential skillset for students as it not only applies to mathematics but also to various fields where logical reasoning is paramount.
Sequence Terms Inequality
Inequalities involving terms of sequences are often vital in establishing relationships between the limits of sequences and other numerical values. Consider a sequence \(a_n\) that is bounded above by a constant \(M\), meaning \(a_n \leq M\) for all \(n\). This inequality sets up a crucial piece of the puzzle when determining the limit of the sequence. Consequently, when we say that a sequence is bounded above by \(M\), we are asserting that \(M\) serves as a ceiling that the sequence terms cannot surpass.

This concept enforces a limit on the growth of the sequence and is essential for students to understand since it directly influences the behavior of the sequence's limit. If a sequence is to converge, then the terms must not only get close to one another but also must not exceed certain bounds鈥攗nderscoring the integral nature of sequence terms' inequalities.
Upper Bound of Sequences
The upper bound of a sequence is a number that no term in the sequence exceeds. Mathematical sequences can have bounds that are either strict or non-strict. For a sequence \(a_n\), if every term is less than or equal to some number \(M\), then \(M\) is an upper bound of the sequence. This concept is key in understanding the limit of a convergent sequence because if a sequence is converging to a limit \(L\) and there is an upper bound \(M\), then \(L\) cannot exceed \(M\).

When solving problems involving the convergence of sequences, recognizing the role of upper bounds helps with visualizing and constraining the behavior of the sequence as it approaches its limit. For students, this context is essential in grasping the broader implications of sequence behavior in calculus and analysis.

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Most popular questions from this chapter

Probability density functions. A nonnegative function \(f\) defined on \((-\infty , \infty)\) is called a probability density function if $$\int_{-\infty}^{\infty} f(x) d x=1.$$ Let \(k>0 .\) Show that the function $$f(x)=\left\\{\begin{aligned}k e^{-k r}, & x \geq 0 \\\0, & x < 0 \end{aligned}\right.$$ is a probability density function. It is called the exponential density function.

You have seen that for all real \(x\).$$\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}$$,However, the rate of convergence is different at different X. Verify that with \(n=100,(1+1 / n) ^ {n}\) is within \(1 \%\) of its \(\begin{array}{ll}\text { (1) } & \text { (i) } 4 \text { is twithin } 1 \% \text { the } \\ \text { the the the the the the the the parition the parition }\end{array}\) limit, while \((1+5 / n)^{n}\) is still about \(12 \%\) from its limit. Give comparable accuracy estimates at \(x=1\) and \(a x=5\) for \(n=1000\).

The differential equation satisfied by the velocity of an object of mass \(m\) dropped from rest under the influence of gravity with air resistance directly proportional to the velocity can be written $$m \frac{d v}{d t}+k v=m g$$ where \(k>0\) is the constant of proportionality, \(g\) is the gravitational constant and \(v(0)=0 .\) The velocity of the object at time \(t\) is given by $$v(t)=(m g / k)\left(1-e^{-(k / m)}\right)$$ (a) Fix \(t\) and find \(\lim _{k \rightarrow 0^{-}} \nu(t)\) (b) Set \(k=0\) in \((*)\) and solve the initial-value problem $$m \frac{d v}{d t}=n t g, \quad v(0)=0$$ Docs this result fit in with what you found in part (a)?

Use comparison test (11.7.2) to determine whether the integral converges. $$\int_{e}^{\infty} \frac{d x}{\sqrt{x+1} \ln x}$$

Find the limit of the sequence. $$\lim _{n \rightarrow \infty} n^{\sin (\pi / n)}$$
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