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Chapter 8: Problem 47

Consider the following sequences. a. Find the first four terms of the sequence. b. Based on part (a) and the figure, determine a plausible limit of the sequence. $$a_{n}=2+2^{-n} ; n=1,2,3, \dots$$

Short Answer

Expert verified
Based on the given sequence $$a_n = 2 + 2^{-n}$$, the first four terms are $$\frac{5}{2}, \frac{9}{4}, \frac{17}{8}, \frac{33}{16}$$. As n approaches infinity, the sequence appears to converge to a limit of 2.

Step by step solution

01

Find the first four terms of the sequence

To calculate the first four terms of the sequence, we need to plug the numbers 1, 2, 3, and 4 into the formula $$a_{n} = 2 + 2^{-n}$$. For n = 1, $$a_{1} = 2 + 2^{-1} = 2 + \frac{1}{2} = \frac{5}{2}$$ For n = 2, $$a_{2} = 2 + 2^{-2} = 2 + \frac{1}{4} = \frac{9}{4}$$ For n = 3, $$a_{3} = 2 + 2^{-3} = 2 + \frac{1}{8} = \frac{17}{8}$$ For n = 4, $$a_{4} = 2 + 2^{-4} = 2 + \frac{1}{16} = \frac{33}{16}$$ So, the first four terms of the sequence are: $$\frac{5}{2}, \frac{9}{4}, \frac{17}{8}, \frac{33}{16}$$.
02

Determine a plausible limit for the sequence

To find the limit of the sequence as n approaches infinity, we need to analyze the behavior of the terms. As n gets larger, the value of $$2^{-n}$$ gets smaller, tending to zero. The overall sequence is generated by adding 2 to these diminishing fractions, so the sequence gradually approaches 2. Therefore, a plausible limit for the sequence as n approaches infinity is 2.

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Key Concepts

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

Limits of Sequences
Understanding the limit of a sequence is crucial when analyzing its long-term behavior. In essence, the limit of a sequence is the value that the terms of the sequence approach as the index, usually denoted by 'n', approaches infinity. A sequence \(a_n\) converges to a limit \(L\) if for every positive number \(蔚>0\), there's a positive integer \(N\) such that for all \(n鈮\), the terms of the sequence \(a_n\) satisfy \( |a_n - L| < 蔚 \).

This might seem abstract, so let's consider the given sequence \(a_n = 2 + 2^{-n}\). As \(n\) becomes very large, the term \(2^{-n}\) gets smaller and smaller, essentially becoming negligible. Therefore, the terms of the sequence get closer and closer to the number 2. In mathematical terms, we express this by saying the limit of the sequence is 2. This notion of a limit is fundamental in calculus and is used to rigorously define concepts like continuity and the convergence of series.
Sequence Terms Calculation
To effectively understand a sequence, it's essential to calculate its individual terms. The terms of a sequence, represented by \(a_n\), are the output of a formula that depends on the position \(n\), within the sequence. For instance, in the given sequence \(a_n = 2 + 2^{-n}\), you can calculate the first four terms by substituting \(n\) with the first four positive integers.

Let's break it down:
  • For \(n = 1\), the term is \(\frac{5}{2}\).
  • For \(n = 2\), it's \(\frac{9}{4}\).
  • \(n = 3\) yields \(\frac{17}{8}\).
  • And \(n = 4\) gives us \(\frac{33}{16}\).
Each term is determined by its position in the sequence and the general form of the sequence itself. It is the evaluation of the sequence's formula at specific points that provides us with the sequence's values.
Infinite Sequences
An infinite sequence continues indefinitely without termination, defined by a formula that can award a specific value to every positive integer. The sequence provided, \(a_n = 2 + 2^{-n}\), is infinite because for every positive integer \(n\), there is a corresponding term \(a_n\).

Studying these sequences involves looking at the behavior of their terms as \(n\) grows without bound. Some infinite sequences converge to a finite limit, while others may diverge, meaning they don't settle on any single value as \(n\) goes to infinity. Convergent sequences like the one we've discussed (\(2 + 2^{-n}\)) are particularly interesting because they offer insight into the values that sequences can stabilize around, providing a sense of 'closeness' to a particular number even as the sequence continues endlessly.

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Most popular questions from this chapter

\(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} \operatorname{In} 1734,\) Leonhard Euler informally proved that \(\sum_{k=1}^{\infty} \frac{1}{k^{2}}=\frac{\pi^{2}}{6} .\) An elegant proof is outlined here that uses the inequality $$\cot ^{2} x<\frac{1}{x^{2}}<1+\cot ^{2} x\left(\text { provided that } 0

Use Theorem 8.6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{e^{n / 10}}{2^{n}}\right\\}$$

Suppose an alternating series \(\sum(-1)^{k} a_{k}\) with terms that are non increasing in magnitude, converges to \(S\) and the sum of the first \(n\) terms of the series is \(S_{n} .\) Suppose also that the difference between the magnitudes of consecutive terms decreases with \(k .\) It can be shown that for \(n \geq 1\) \(\left|S-\left(S_{n}+\frac{(-1)^{n+1} a_{n+1}}{2}\right)\right| \leq \frac{1}{2}\left|a_{n+1}-a_{n+2}\right|\) a. Interpret this inequality and explain why it is a better approximation to \(S\) than \(S_{n}\) b. For the following series, determine how many terms of the series are needed to approximate its exact value with an error less than \(10^{-6}\) using both \(S_{n}\) and the method explained in part (a). (i) \(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k}\) (ii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{k \ln k}\) (iii) \(\sum_{k=2}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\)

Here is a fascinating (unsolved) problem known as the hailstone problem (or the Ulam Conjecture or the Collatz Conjecture). It involves sequences in two different ways. First, choose a positive integer \(N\) and call it \(a_{0} .\) This is the seed of a sequence. The rest of the sequence is generated as follows: For \(n=0,1,2, \ldots\) $$a_{n+1}=\left\\{\begin{array}{ll} a_{n} / 2 & \text { if } a_{n} \text { is even } \\ 3 a_{n}+1 & \text { if } a_{n} \text { is odd .} \end{array}\right.$$ However, if \(a_{n}=1\) for any \(n,\) then the sequence terminates. a. Compute the sequence that results from the seeds \(N=2,3\), \(4, \ldots, 10 .\) You should verify that in all these cases, the sequence eventually terminates. The hailstone conjecture (still unproved) states that for all positive integers \(N\), the sequence terminates after a finite number of terms. b. Now define the hailstone sequence \(\left\\{H_{k}\right\\},\) which is the number of terms needed for the sequence \(\left\\{a_{n}\right\\}\) to terminate starting with a seed of \(k\). Verify that \(H_{2}=1, H_{3}=7\), and \(H_{4}=2\). c. Plot as many terms of the hailstone sequence as is feasible. How did the sequence get its name? Does the conjecture appear to be true?

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=e^{n / 2} \text { and } b_{n}=n^{5}, n \geq 2$$
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