/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Write the terms \(a_{1}, a_{2}, ... [FREE SOLUTION] | 91Ó°ÊÓ

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Write the terms \(a_{1}, a_{2}, a_{3},\) and \(a_{4}\) of the following sequences. If the sequence appears to converge, make a conjecture about its limit. If the sequence diverges, explain why. $$a_{n}=1-10^{-n} ; n=1,2,3, \ldots$$

Short Answer

Expert verified
Based on the given sequence \(a_{n} = 1 - 10^{-n}\), the first four terms of the sequence are \(a_{1} = \frac{9}{10},\ a_{2} = \frac{99}{100},\ a_{3} = \frac{999}{1000},\) and \(a_{4} = \frac{9999}{10000}\). As n increases, the terms of the sequence get closer to 1. The sequence converges to the limit 1 as \(n\rightarrow\infty\).

Step by step solution

01

Calculate the terms a1, a2, a3, and a4

We can find the terms of the sequence by substituting the values of n into the formula \(a_{n} = 1 - 10^{-n}\). For \(n = 1\): $$a_{1} = 1 - 10^{-1} = 1 - \frac{1}{10} = \frac{9}{10}$$ For \(n = 2\): $$a_{2} = 1 - 10^{-2} = 1 - \frac{1}{100} = \frac{99}{100}$$ For \(n = 3\): $$a_{3} = 1 - 10^{-3} = 1 - \frac{1}{1000} = \frac{999}{1000}$$ For \(n = 4\): $$a_{4} = 1 - 10^{-4} = 1 - \frac{1}{10000} = \frac{9999}{10000}$$
02

Determine if the sequence converges or diverges

Observing the values of the terms \(a_{1}, a_{2}, a_{3},\) and \(a_{4}\), we can notice that their values are getting closer and closer to 1 as n increases. In the case of this sequence, when \(n\rightarrow\infty\), the term \(10^{-n}\) approaches 0, since for a large value of \(n\), the denominator (10^n) will be a very large number and thus the fraction will be very small. Therefore, the limit of the sequence is given by: $$\lim_{n\rightarrow\infty}(1 - 10^{-n}) = 1 - 0 = 1$$ As we can see that the sequence converges to the limit 1 for larger values of n. The terms of the sequence are: $$a_{1} = \frac{9}{10},\ a_{2} = \frac{99}{100},\ a_{3} = \frac{999}{1000},\ a_{4} = \frac{9999}{10000},\ \text{and the limit is}\ 1$$

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Most popular questions from this chapter

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0}\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G\). a. Show that \(a_{n}>b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

Given any infinite series \(\Sigma a_{k},\) let \(N(r)\) be the number of terms of the series that must be summed to guarantee that the remainder is less than \(10^{-r}\), where \(r\) is a positive integer. a. Graph the function \(N(r)\) for the three alternating \(p\) -series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{p}},\) for \(p=1,2,\) and \(3 .\) Compare the three graphs and discuss what they mean about the rates of convergence of the three series. b. Carry out the procedure of part (a) for the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k !}\) and compare the rates of convergence of all four series.

Series of squares Prove that if \(\sum a_{k}\) is a convergent series of positive terms, then the series \(\Sigma a_{k}^{2}\) also converges.

Evaluate the limit of the following sequences. $$a_{n}=\frac{n^{8}+n^{7}}{n^{7}+n^{8} \ln n}$$

Evaluate the limit of the following sequences. $$a_{n}=\cos \left(0.99^{n}\right)+\frac{7^{n}+9^{n}}{63^{n}}$$

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