/*! 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 103 Prove, using the definition of t... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 103

Prove, using the definition of the limit of a sequence, that \(\lim _{n \rightarrow \infty} r^{n}=0\) for \(-1

Short Answer

Expert verified
To prove \( \lim_{n \to \infty} r^n = 0 \) for \( -1 < r < 1 \), we use the definition of a limit and the properties of logarithms to show that there exists an integer \( N \) such that for all \( n > N \), \( |r^n| < \epsilon \).

Step by step solution

01

- Define the Problem in Terms of the Limit Definition

The definition of a sequence limit \( \lim_{n \to \infty} a_n = L \) is: given any small number \( \epsilon > 0 \), there is an integer \( N \) such that whenever \( n > N \), then \( |a_n - L| < \epsilon \). We must show this for \( a_n = r^n \) and \( L = 0 \). So our task is to find \( N \) such that if \( n > N \) then \( |r^n - 0| < \epsilon \), or equivalently \( |r^n| < \epsilon \), for a given \( \epsilon > 0 \).
02

- Find an Appropriate Value for N

We know that \( -1 < r < 1 \), so \( |r| < 1 \). We want to find \( N \) such that \( |r^n| < \epsilon \). As \( |r| < 1 \), \( |r^n| \) gets smaller as \( n \) gets bigger, so we can make \( |r^n| < \epsilon \) by choosing \( n \) big enough. More specifically, we can use the properties of logarithms to show that if \( n > \frac{ \log(\epsilon) }{ \log(|r|) } \), then \( |r^n| < \epsilon \).
03

- Conclusion

Therefore, by choosing any \( N \) greater than \( \frac{ \log(\epsilon) }{ \log(|r|) }\), we can ensure that \( |r^n - 0| = |r^n| < \epsilon \) for all \( n > N \). This satisfies the definition of the limit, so we can conclude that \( \lim_{n \to \infty} r^n = 0 \) for \( -1 < r < 1 \).

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(n>1\), then \(n !=n(n-1) !\)

(a) Show that \(\int_{1}^{n} \ln x d x<\ln (n !)\) for \(n \geq 2\). (b) Draw a graph similar to the one above that shows \(\ln (n !)<\int_{1}^{n+1} \ln x d x\) (c) Use the results of parts (a) and (b) to show that \(\frac{n^{n}}{e^{n-1}}1\) (d) Use the Squeeze Theorem for Sequences and the result of part (c) to show that \(\lim _{n \rightarrow \infty} \frac{\sqrt[n]{n !}}{n}=\frac{1}{e}\) (e) Test the result of part (d) for \(n=20,50,\) and 100 .

In an experiment, three people toss a fair coin one at a time until one of them tosses a head. Determine, for each person, the probability that he or she tosses the first head. Verify that the sum of the three probabilities is 1 .

A fair coin is tossed repeatedly. The probability that the first head occurs on the \(n\) th toss is given by \(P(n)=\left(\frac{1}{2}\right)^{n},\) where \(n \geq 1\) (a) Show that \(\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n}=1\). (b) The expected number of tosses required until the first head occurs in the experiment is given by \(\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}\) Is this series geometric? (c) Use a computer algebra system to find the sum in part (b).

Use the formula for the \(n\) th partial sum of a geometric series $$\sum_{i=0}^{n-1} a r^{i}=\frac{a\left(1-r^{n}\right)}{1-r}$$ You go to work at a company that pays \(\$ 0.01\) for the first day, \(\$ 0.02\) for the second day, \(\$ 0.04\) for the third day, and so on. If the daily wage keeps doubling, what would your total income be for working (a) 29 days, (b) 30 days, and (c) 31 days?
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