Question

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

Question: Prove that the limit of the sequence \(a_n = \frac{1}{n^2}\) as \(n\) approaches infinity is equal to 0 using the formal definition of the limit of a sequence. Short Answer: Find an integer \(N\) such that for all \(n > N\), the inequality \(\frac{1}{n^2} < \epsilon\) holds, where \(\epsilon > 0\). We can show that by choosing \(N = \lceil \sqrt{\frac{1}{\epsilon}} \rceil\), we satisfy the condition for any given \(\epsilon > 0\). Therefore, the limit of the sequence \(a_n = \frac{1}{n^2}\) as \(n\) approaches infinity is equal to 0.

Step-by-step solution

Set up the inequality

We want to find an integer \(N\) such that for all \(n > N\), the inequality \(\frac{1}{n^2} < \epsilon\) holds, where \(\epsilon > 0\).

Manipulate the inequality

First, we want to isolate \(n\) in the inequality. Start by taking the reciprocal of both sides: $$n^2 > \frac{1}{\epsilon}$$ Now, take the square root of both sides: $$n > \sqrt{\frac{1}{\epsilon}}$$

Find an integer N

We can find such an integer \(N\) by taking the ceiling of \(\sqrt{\frac{1}{\epsilon}}\). The ceiling function is denoted \(\lceil x \rceil\) and is defined as the smallest integer greater than or equal to \(x\). So, $$N = \lceil \sqrt{\frac{1}{\epsilon}} \rceil$$

Verify that N satisfies the definition of the limit

We now need to show that for all \(n > N\), the inequality \(\frac{1}{n^2} < \epsilon\) holds. Since \(n > N \geq \sqrt{\frac{1}{\epsilon}}\), we can deduce that \(n^2 > \frac{1}{\epsilon}\). Thus, \(\frac{1}{n^2} < \epsilon\) and our choice of \(N\) satisfies the condition.

Conclusion

We have shown that for any given \(\epsilon > 0\), there exists an integer \(N\) such that for all \(n > N\), the inequality \(\frac{1}{n^2} < \epsilon\) holds. Using the formal definition of the limit of a sequence, this implies that $$\lim _{n \rightarrow \infty} \frac{1}{n^{2}}=0$$