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Chapter 11: Problem 77

Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(23-84\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\left(\frac{1}{3}\right)^{n}+\frac{1}{\sqrt{2^{n}}} $$

Short Answer

Expert verified
Sequence \(a_n\) converges to 0.

Step by step solution

01

Analyze Each Term in the Sequence

The sequence given is \(a_n = \left(\frac{1}{3}\right)^n + \frac{1}{\sqrt{2^n}}\). The sequence consists of two terms: the first term is \(\left(\frac{1}{3}\right)^n\) and the second is \(\frac{1}{\sqrt{2^n}}\). We will analyze the limit of each term separately as \(n\) approaches infinity.
02

Understand the First Term

Consider the first term \(\left(\frac{1}{3}\right)^n\). As \(n\) increases, \(\left(\frac{1}{3}\right)^n\) becomes smaller and approaches 0 because the fraction \(\frac{1}{3}\) is less than 1 and raising it to a higher power makes it closer to zero. Thus, \(\lim_{n \to \infty}\left(\frac{1}{3}\right)^n = 0\).
03

Understand the Second Term

Now, consider the second term \(\frac{1}{\sqrt{2^n}}\). This can be rewritten as \(2^{-n/2}\). As \(n\) increases, \(2^{-n/2}\) also becomes smaller and approaches 0 because \(2^n\) grows exponentially making \(\frac{1}{\sqrt{2^n}}\) decrease. Therefore, \(\lim_{n \to \infty}\frac{1}{\sqrt{2^n}} = 0\).
04

Combine the Limits

Since both terms of the sequence approach 0 as \(n\) becomes large, the entire sequence \(a_n\) also approaches 0. Therefore, the sum of the limits is 0: \(\lim_{n \to \infty}a_n = 0\). This means the sequence converges to this limit.

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Key Concepts

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

Limit of a Sequence
A core idea in understanding sequences is the concept of a limit. When we talk about the limit of a sequence, we're looking at what value the sequence approaches as the number of terms goes to infinity. For example, consider a sequence like \(a_n = (\frac{1}{3})^n + \frac{1}{\sqrt{2^n}}\). To determine if it converges, we check if it approaches a specific number as \(n\) goes to infinity.

In simple terms, think of a sequence as a list of numbers. If the numbers in this list keep getting closer and closer to some particular value, then the sequence converges on that limit. In the example above, as \(n\) gets larger, \(\left(\frac{1}{3}\right)^n\) and \(\frac{1}{\sqrt{2^n}}\) approach zero. Thus, the entire sequence moves towards zero, which makes 0 the limit of this sequence. Understanding whether sequences converge or not, and finding their limits, reveals the behavior of sequences over the long term.
Exponential Decay
Exponential decay occurs when a quantity decreases at a rate proportional to its current value. When examining the first term of our sequence, \(\left(\frac{1}{3}\right)^n\), we observe exponential decay. Here, the base \(\frac{1}{3}\) is less than one, so raising it to the \(n\)-th power causes the term to shrink rapidly towards zero as \(n\) increases.

This decay happens because with each increase in \(n\), the fraction \(\frac{1}{3}\) is multiplied by itself, producing smaller and smaller values. You can visualize this as a light dimmer switch that progressively reduces the brightness of a lamp, where each twist makes the light even fainter. As such, determining exponential decay involves identifying sequences where terms decrease by a consistent factor, quickly approaching a value near zero.
Infinite Series
An infinite series is the sum of the terms of an infinite sequence. The significance of an infinite series lies in determining whether its sum approaches a finite number as the number of terms increases without bound. Each term in a converging infinite series contributes to making the total sum approach a specific limit.

In the case of our sequence \(a_n = \left(\frac{1}{3}\right)^n + \frac{1}{\sqrt{2^n}}\), imagine adding up each term one by one. Even though there are infinitely many terms, if their cumulative sum approaches a particular number, the infinite series converges. Conversely, if the sum diverges, it doesn't approach any specific value.
  • Converging series: These have a limit, meaning they stabilize to a particular sum.
  • Diverging series: These lack a limit and their sum can grow infinitely large or oscillate indefinitely.
The sum of \(\left(\frac{1}{3}\right)^n\) and \(\frac{1}{\sqrt{2^n}}\) eventually zeros out in an infinite series context, confirming convergence to zero.

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Most popular questions from this chapter

Outline of the proof of the Rearrangement Theorem a. Let \(\epsilon\) be a positive real number, let \(L=\sum_{n=1}^{\infty} a_{n}\), and let \(s_{k}=\sum_{n=1}^{k} a_{n} .\) Show that for some index \(N_{1}\) and for some index \(N_{2} \geq N_{1}\), $$ \sum_{n=N_{1}}^{\infty}\left|a_{n}\right|<\frac{\epsilon}{2} \text { and }\left|s_{N_{2}}-L\right|<\frac{\epsilon}{2} $$ Since all the terms \(a_{1}, a_{2}, \ldots, a_{N_{2}}\) appear somewhere in the sequence \(\left\\{b_{n}\right\\}\), there is an index \(N_{3} \geq N_{2}\) such that if \(n \geq N_{3}\), then \(\left(\sum_{k=1}^{n} b_{k}\right)-s_{N_{2}}\) is at most a sum of terms \(a_{m}\) with \(m \geq N_{1}\). Therefore, if \(n \geq N_{3}\), $$ \begin{aligned} \left|\sum_{k=1}^{n} b_{k}-L\right| & \leq\left|\sum_{k=1}^{n} b_{k}-s_{N_{2}}\right|+\left|s_{N_{2}}-L\right| \\ & \leq \sum_{k=N_{1}}^{\infty}\left|a_{k}\right|+\left|s_{N_{2}}-L\right|<\epsilon \end{aligned} $$ b. The argument in part (a) shows that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely then \(\sum_{n=1}^{\infty} b_{n}\) converges and \(\sum_{n=1}^{\infty} b_{n}=\sum_{n=1}^{\infty} a_{n}\) Now show that because \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, \(\sum_{n=1}^{\infty}\left|b_{n}\right|\) converges to \(\sum_{n=1}^{\infty}\left|a_{n}\right|\).

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{y \rightarrow 0} \frac{\tan ^{-1} y-\sin y}{y^{3} \cos y} $$

Prove that if \(\left\\{a_{n}\right\\}\) is a convergent sequence, then to every positive number \(\epsilon\) there corresponds an integer \(N\) such that for all \(m\) and \(n\) , $$ m>N \text { and } n>N \Rightarrow\left|a_{m}-a_{n}\right|<\epsilon $$

Series for tan \(^{-1} x\) for \(|x|>1\) Derive the series $$ \begin{array}{l}{\tan ^{-1} x=\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x>1} \\ {\tan ^{-1} x=-\frac{\pi}{2}-\frac{1}{x}+\frac{1}{3 x^{3}}-\frac{1}{5 x^{5}}+\cdots, \quad x<-1}\end{array} $$ by integrating the series $$ \frac{1}{1+t^{2}}=\frac{1}{t^{2}} \cdot \frac{1}{1+\left(1 / t^{2}\right)}=\frac{1}{t^{2}}-\frac{1}{t^{4}}+\frac{1}{t^{6}}-\frac{1}{t^{8}}+\cdots $$ in the first case from \(x\) to \(\infty\) and in the second case from \(-\infty\) to \(x .\)

For what values of \(a,\) if any, do the series in converge? $$\sum_{n=3}^{\infty}\left(\frac{1}{n-1}-\frac{2 a}{n+1}\right)$$
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