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Use the Squeeze Theorem and Corollary 1.1 to prove that the sequence converges to 0 (given that $$.\lim _{n \rightarrow \infty} \frac{1}{n}=\lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0)$$. $$a_{n}=(-1)^{n} \frac{\ln n}{n^{2}}$$

Short Answer

Expert verified
\(\lim_{n \to \infty} (-1)^n \frac{\ln n}{n^2} = 0\).

Step by step solution

01

Identify the sequence and establish boundaries

Take the sequence \(a_n=(-1)^n \frac{\ln n}{n^2}\) and establish a lower and upper boundary. In this case, \(-\frac{\ln n}{n^2}\) and \(\frac{\ln n}{n^2}\). Since \((-1)^n\) oscilates between -1 and 1, we can make this assumption.
02

Evaluate the limit of the boundaries

We have established that \(\lim_{n \to \infty} \frac{1}{n} = \lim_{n \to \infty} \frac{1}{n^2} = 0\), so we can now evaluate the limit of our sequence boundaries. As \(n \to \infty\), \(\frac{\ln n}{n^2}\) will approach 0 (due to the existing limits).
03

Apply the Squeeze Theorem

By Squeeze theorem, if \(\lim_{n \to \infty} b_n = \lim_{n \to \infty} c_n = L\), and \(b_n \leq a_n \leq c_n\) for all \(n \geq k\), then \(\lim_{n \to \infty} a_n = L\). In this case, we established \(b_n = \lim_{n \to \infty} -\frac{\ln n}{n^2}\) and \(c_n = \lim_{n \to \infty} \frac{\ln n}{n^2}\) which both tend to 0. We also established that \(-\frac{\ln n}{n^2} \leq (-1)^n \frac{\ln n}{n^2} \leq \frac{\ln n}{n^2}\). Thus, by the Squeeze Theorem, \(\lim_{n \to \infty} (-1)^n \frac{\ln n}{n^2} = 0\).

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Key Concepts

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

Convergent Sequences
A convergent sequence is a sequence of numbers that approaches a single, finite number as the number of terms grows indefinitely. In other words, as you progress further and further into the sequence, the terms become closer and closer to a specific value, which is known as the limit of the sequence. For example, the sequence \( \frac{1}{n} \) converges to 0, as no matter how large \( n \) becomes, \( \frac{1}{n} \) will get smaller and smaller, approaching, but never actually reaching, zero.

Understanding convergence is essential in calculus because it helps us analyse the behaviour of sequences and functions at infinity. To determine whether a sequence \( a_n \) is convergent, we typically try to look for a pattern or apply known theorems, such as the Squeeze Theorem, to establish the limit. In our exercise example, showing that \( a_n = (-1)^n \frac{\ln n}{n^2} \) converges involves the idea that this sequence gets trapped between two convergent sequences that both approach zero, hence \( a_n \) must also approach zero.
Limits of Sequences
The concept of limits is fundamental in understanding how sequences behave as they progress towards infinity. In calculus, the limit of a sequence \( a_n \) as \( n \) approaches infinity is denoted as \( \lim_{n \to \infty} a_n \). This notation is used to express the value that the terms of the sequence get arbitrarily close to, assuming it exists.

To calculate the limit of a sequence, we may use direct substitution, algebraic manipulations, or apply special theorems if the sequence is more complex. In the exercise, the limits of \( \frac{1}{n} \) and \( \frac{1}{n^2} \) as \( n \) goes to infinity were both used to establish that no matter how large \( n \) gets, these expressions become insignificantly small, contributing to the idea that our sequence \( a_n \) will have a boundary.
Calculus
Calculus is a branch of mathematics that deals with rates of change (differential calculus) and accumulation of quantities (integral calculus). It provides powerful tools, such as the concept of a limit, to analyze changes and motion, and to solve problems involving complex sequences and functions.

The Squeeze Theorem used in our exercise illustrates an application of calculus to sequence analysis. This theorem allows us to conclude information about a complex sequence by comparing it to simpler ones we understand well. In implementing calculus to our problem, we didn't calculate the limit directly; instead, we bounded it between two other sequences that converge to the same limit, demonstrating that the limits in calculus often require creative approaches and a deep understanding of the behaviours of mathematical expressions as they stretch toward infinity.

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Most popular questions from this chapter

Find the interval of convergence of the (nonpower) series and the corresponding series of derivatives. $$\sum_{k=1}^{\infty} \frac{\cos (x / k)}{k}$$

If \(\sum_{k=0}^{\infty} a_{k} x^{k}\) has radius of convergence \(r,\) with \(0 < r < \infty,\) determine the radius of convergence of \(\sum_{k=0}^{\infty} a_{k}\left(\frac{x}{b}\right)^{k}\) for any constant \(b \neq 0\)

Find the interval of convergence of the (nonpower) series and the corresponding series of derivatives. $$\sum_{k=1}^{\infty} \frac{\cos \left(k^{3} x\right)}{k^{2}}$$

This problem is sometimes called the coupon collectors" problem. The problem is faced by collectors of trading cards. If there are \(n\) different cards that make a complete set and you randomly obtain one at a time, how many cards would you expect to obtain before having a complete set? (By random, we mean that each different card has the same probability of \(\frac{1}{n}\) of being the next card obtained.) In exercises \(69-72,\) we find the answer for \(n=10 .\) The first step is simple; to collect one card you need to obtain one card. Now, given that you have one card, how many cards do you need to obtain to get a second (different) card? If you're lucky, the next card is it (this has probability 10). But your next card might be a duplicate, then you get a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{9}{10}\right) .\) Or you might get two duplicates and then a new card (this has probability \(\left.\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}\right) ;\) and so on. The mean is \(1 \cdot \frac{9}{10}+2 \cdot \frac{1}{10} \cdot \frac{9}{10}+3 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{9}{10}+\cdots\) or \(\sum_{k=1}^{\infty} k\left(\frac{1}{10}\right)^{k-1}\left(\frac{9}{10}\right)=\sum_{k=1}^{\infty} \frac{9 k}{10^{k}} .\) Using the same trick as in exercise \(68,\) show that this is a convergent series with \(\operatorname{sum} \frac{10}{9}\)

Use the Binomial Theorem to find the first five terms of the Maclaurin series. $$f(x)=\left(1+x^{2}\right)^{4 / 5}$$

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