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Chapter 9: Problem 70

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0$$

Short Answer

Expert verified
Question: Prove that the limit of the sequence \(a_n = \frac{1}{n^2}\) is 0 using the formal definition of the limit of a sequence. Answer: For any \(\epsilon > 0\), we found a positive integer \(N = \lceil\sqrt{\frac{1}{\epsilon}}\rceil\). We showed that for any integer \(n \geq N\), it follows that \(\frac{1}{n^2} < \epsilon\). Therefore, by the formal definition of the limit, we have proven that \(\lim_{n \rightarrow \infty} \frac{1}{n^2} = 0\).

Step by step solution

01

Use the formal definition of the limit of a sequence

We need to show that for any \(\epsilon > 0\), there exists a positive integer \(N\) such that for any integer \(n \geq N\), it follows that \(|a_n - L| < \epsilon\). In this case, \(a_n = \frac{1}{n^2}\) and \(L = 0\).
02

Simplify the inequality

We want to show that for any \(\epsilon > 0\), there exists a positive integer \(N\) such that for any integer \(n \geq N\), it follows that \(|\frac{1}{n^2} - 0| < \epsilon\). Since \(|x|\) is always non-negative, we need to show that \(\frac{1}{n^2} < \epsilon\).
03

Manipulate the inequality to find \(N\)

To find \(N\), we can change the inequality \(\frac{1}{n^2} < \epsilon\) to \(n^2 > \frac{1}{\epsilon}\) by taking the reciprocal of both sides. Then, we can take the square root of both sides to get \(n > \sqrt{\frac{1}{\epsilon}}\). Since \(n\) must be an integer, we can find a value for \(N\) by taking the smallest integer greater than \(\sqrt{\frac{1}{\epsilon}}\). This gives us \(N = \lceil\sqrt{\frac{1}{\epsilon}}\rceil\).
04

Prove the inequality holds for \(n \geq N\)

We found that \(N = \lceil\sqrt{\frac{1}{\epsilon}}\rceil\). Now, we need to show that the inequality \(\frac{1}{n^2} < \epsilon\) holds for any integer \(n\) such that \(n \geq N\). Since \(n \geq N = \lceil\sqrt{\frac{1}{\epsilon}}\rceil \geq \sqrt{\frac{1}{\epsilon}}\), we can see that \(n^2 \geq \frac{1}{\epsilon}\). Thus, \(\frac{1}{n^2} \leq \epsilon\). Since the division is strictly decreasing function in the positive numbers domain, the inequality becomes strict, so we have \(\frac{1}{n^2} < \epsilon\).
05

Conclusion

Using the formal definition of the limit of a sequence and the manipulation of inequalities, we proved that \(\lim_{n \rightarrow \infty} \frac{1}{n^2} = 0\).

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Most popular questions from this chapter

For what values of \(x\) does the geometric series $$f(x)=\sum_{k=0}^{\infty}\left(\frac{1}{1+x}\right)^{k}$$ converge? Solve \(f(x)=3\)

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