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Chapter 9: Problem 69

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{1}{n}=0$$

Short Answer

Expert verified
Question: Prove that the limit of the sequence \(\frac{1}{n}\) as \(n\) approaches infinity is 0 using the formal definition of the limit of a sequence. Answer: By the formal definition of the limit of a sequence, a sequence \((a_n)\) converges to the limit \(L\) if for any given \(\epsilon > 0\), there exists a positive integer \(N\) such that for all \(n > N\), we have \(|a_n - L| < \epsilon\). Applying this definition to the sequence \(\frac{1}{n}\), we find an integer \(N = \left\lceil\frac{1}{\epsilon}\right\rceil\) such that for all \(n > N\), the inequality \(\frac{1}{n} < \epsilon\) holds true. This satisfies the definition and proves that \(\lim _{n \rightarrow \infty} \frac{1}{n}=0\).

Step by step solution

01

Write down the formal definition of the limit of a sequence

The formal definition of the limit of a sequence states that a sequence \((a_n)\), with \(n\) as a positive integer, converges to the limit \(L\) if, for any given \(\epsilon > 0\), there exists a positive integer \(N\) such that for all \(n > N\), we have: $$|a_n - L| < \epsilon$$ In this problem, we must investigate if the given sequence \(\frac{1}{n}\) converges to the limit 0 as \(n\) approaches infinity.
02

Write the expression for the absolute value of the difference between the sequence and its limit

The next step is to write down the absolute value of the difference between the terms of the sequence \(\frac{1}{n}\) and the claimed limit 0. This expression takes the form: $$\left|\frac{1}{n} - 0\right|$$ which simplifies to: $$\frac{1}{n}$$
03

Find an integer \(N\) so that the expression is less than \(\epsilon\)

To confirm the limit is indeed 0 using the definition, we must find an integer \(N\) such that for all \(n > N\), the inequality: $$\frac{1}{n} < \epsilon$$ holds true. By simple algebra, this inequality is equivalent to: $$n > \frac{1}{\epsilon}$$ We can choose \(N\) to be the smallest integer greater than \(\frac{1}{\epsilon}\), which can be written as: $$N = \left\lceil\frac{1}{\epsilon}\right\rceil$$
04

Prove that it works for all \(n > N\)

Now, we must prove that our choice of \(N\) guarantees that \(\frac{1}{n} < \epsilon\) for all \(n > N\). Since \(n > N \geq \frac{1}{\epsilon}\), it follows that: $$\frac{1}{n} \leq \frac{1}{N} < \frac{1}{\frac{1}{\epsilon}} = \epsilon$$ Thus, we have shown that for all \(n > N\), the inequality \(\frac{1}{n} < \epsilon\) holds true. This satisfies the formal definition of the limit of a sequence, and hence, we can conclude that: $$\lim _{n \rightarrow \infty} \frac{1}{n}=0$$

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