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Chapter 9: Problem 72

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} b^{-n}=0, \text { for } b > 1$$

Short Answer

Expert verified
Question: Prove that the limit of the sequence \(b^{-n}\) is 0 as \(n\) approaches infinity for every \(b > 1\). Answer: To prove this, we used the formal definition of the limit of a sequence and showed that for a given \(\epsilon > 0\), there exists an integer \(N\) such that \(|\frac{1}{b^n} - 0| < \epsilon\) for all \(n > N\). By taking the logarithm with base \(b\) and applying its property, we found that \(N = \left\lceil \log_b \frac{1}{\epsilon} \right\rceil\). Hence, we concluded that \(\lim _{n \rightarrow \infty} b^{-n} = 0\) for every \(b > 1\).

Step by step solution

01

Rewrite the sequence

Rewrite the sequence \(b^{-n}\) as \(\frac{1}{b^n}\) to work with positive exponentials. $$b^{-n} = \frac{1}{b^n}$$
02

Define the limit and epsilon

For the given sequence \(\frac{1}{b^n}\), we want to prove that \(\lim _{n \rightarrow \infty} \frac{1}{b^n} = 0\). Let \(\epsilon > 0\) be an arbitrary positive number.
03

Define the inequality and find N

We want to find an integer \(N\) such that the inequality \(|a_n - 0| < \epsilon\) holds for every \(n > N\). In this case, we have: $$\left|\frac{1}{b^n} - 0\right| < \epsilon$$ $$\frac{1}{b^n} < \epsilon$$ We need to find an integer value of \(N\) such that the above inequality holds for all \(n > N\).
04

Use the property of logarithm

We can rewrite the inequality using the logarithm with base \(b\) and apply the property that \(x > y\) is equivalent to \(\log_b x > \log_b y\) if \(b > 1\). $$n\log_b b > \log_b \frac{1}{\epsilon}$$ $$n > \log_b \frac{1}{\epsilon}$$
05

Define N

Since we need an integer value for \(N\), we can take the smallest integer greater than \(\log_b \frac{1}{\epsilon}\). Define: $$N = \left\lceil \log_b \frac{1}{\epsilon} \right\rceil$$
06

Determine the result

With this definition of \(N\), we have: $$\frac{1}{b^n} < \epsilon$$ for all \(n > N\), which means: $$\lim _{n \rightarrow \infty} \frac{1}{b^n} = 0$$ Thus, we proved that for every \(b>1\), $$\lim _{n \rightarrow \infty} b^{-n} = 0$$

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