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Chapter 11: Problem 67

Evaluate numerically the limit of each sequence as \(n \rightarrow \infty\). Some of these sequences converge more rapidly than others. Determine for each sequence the least value of \(n\) for which the \(n\)th term differs from the limit by less than 0.001. $$\frac{1}{n^{2}}$$

Short Answer

Expert verified
The limit of the sequence as \(n\) approaches infinity is 0. The least value of \(n\) for which the \(n\)th term differs from the limit by less than 0.001 is \(n = 32\).

Step by step solution

01

Evaluate the limit of the sequence

To find the limit of the sequence as n approaches infinity, we have: $$\lim_{n \rightarrow \infty} \frac{1}{n^2}$$ Since the denominator goes to infinity, the limit becomes: $$\lim_{n \rightarrow \infty} \frac{1}{n^2} = 0$$
02

Determine the least value of n

We need to find the least value of n for which the nth term differs from the limit (0) by less than 0.001. Mathematically, this can be represented as follows: $$\left| \frac{1}{n^2} - 0 \right| < 0.001$$. Which simplifies to: $$\frac{1}{n^2} < 0.001$$. To solve for n, we need to get rid of the denominator. Multiply both sides by \(n^2\): $$1 < 0.001n^2$$. Divide both sides by 0.001: $$\frac{1}{0.001} < n^2$$. Now take the square root of both sides: $$\sqrt{\frac{1}{0.001}} < n$$ $$\sqrt{1000} < n$$. Since n is an integer value, we must find the smallest integer greater than \(\sqrt{1000}\): $$\lceil \sqrt{1000} \rceil = 32$$. So, the least value of n for which the nth term differs from the limit by less than 0.001 is \(n = 32\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit of a Sequence
In calculus, the concept of the "limit of a sequence" refers to what a sequence approaches as the index goes to infinity. To analyze the behavior of the sequence \( \frac{1}{n^2} \), we look at what happens to this expression as \( n \) becomes very large.

This sequence consists of the terms \( \frac{1}{1}, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \ldots \), and visibly, these values decrease and move towards zero.

When solving the limit mathematically, you write it as:
  • \( \lim_{n \rightarrow \infty} \frac{1}{n^2} \)

Because \( n^2 \) becomes extremely large as \( n \) increases, the fraction itself becomes smaller and smaller. It eventually approaches 0, which signifies that the sequence's limit is zero.

This concept is foundational for understanding how functions and sequences behave at the edge of infinity.
Numerical Evaluation
Numerical evaluation helps to determine at what point a sequence meets certain criteria, like a specific closeness to its limit.

In our example with the sequence \( \frac{1}{n^2} \), we know it approaches a limit of 0.
Now, numerically, we want to find the smallest \( n \) such that \( \frac{1}{n^2} \) is less than the tolerance level we set, which is 0.001 for this exercise.

To do this, we set up the inequality:
  • \( \frac{1}{n^2} < 0.001 \)

Simplifying gives \( n^2 > 1000 \). Thus, by taking the square root, we find \( n > \sqrt{1000} \). Consequently, the smallest integer \( n \) that satisfies this is 32.

Numerical evaluation is essential for making practical decisions when exact calculations can become complex or infeasible.
Convergence of Sequences
The concept of "convergence of sequences" is about whether a sequence approaches a specific number as \( n \) increases indefinitely.

For the sequence \( \frac{1}{n^2} \), we investigated its limit and recognized that its terms get closer to zero as \( n \) grows.

If a sequence converges, it means there is a fixed, finite value that it gets arbitrarily close to, for any small positive distance from that value 鈥� in this case, zero.

To determine convergence, you can set up a test where you find \( n \) such that all terms \( a_n \) for \( n \geq N \) are within a certain distance \( \epsilon \) from the limit (here, \( \epsilon = 0.001 \)). This ensures that \( \frac{1}{n^2} \) is less than \( 0.001 \) beyond some \( n \), and we found that \( n = 32 \) fulfills this requirement. Convergence is a powerful concept that underpins many aspects of calculus and advanced mathematical analysis.

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Most popular questions from this chapter

Evaluate $$\int_{0}^{1} \arcsin x d x$$ using integration by parts even though the technique leads to an improper integral.

(Arithmetic means) For a sequence \(a_{1} \cdot a_{2}, \cdots,\) set $$m_{n}=\frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)$$, (a) Prove that if the \(a_{n}\) form an increasing sequence, then the \(m_{n}\) form an increasing sequence. (b) Prove that if \(a_{n} \rightarrow 0\), then \(m_{x} \rightarrow 0\).

The set \(S\) of rational numbers \(x\) with \(x^{2}<2\) has rational upper bounds but no least rational upper bound. The argument goes like this. Suppose that \(S\) has a least rational upper bound and call it \(x_{0} .\) Then either $$x_{0}^{2}=2, \quad \text { or } \quad x_{0}^{2}>2, \quad \text { or } \quad x_{0}^{2} < 2.$$ (a) Show that \(x_{0}^{2}=2\) is impossible by showing that if \(x_{0}^{2}=\) 2, then \(x_{0}\) is not rational. (b) Show that \(x_{0}^{2}>2\) is impossible by showing that if \(x_{0}^{2}>2,\) then there is a positive integer \(n\) for which \(\left(x_{0}-\frac{1}{n}\right)^{2}>2,\) which makes \(x_{0}-\frac{1}{n}\) a rational upper bound for \(S\) that is less than the least rational upper bound \(x_{0}\). (c) Show that \(x_{0}^{2}<2\) is impossible by showing that if \(x_{0}^{2}<2,\) then there is a positive integer \(n\) for which \(\left(x_{0}+\frac{1}{n}\right)^{2}<2 .\) This places \(x_{0}+\frac{1}{n}\) in \(S\) and show's that \(x_{0}\) cann0ot be an upper bound for \(S\).

You have seen that for all real \(x\).$$\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}$$,However, the rate of convergence is different at different X. Verify that with \(n=100,(1+1 / n) ^ {n}\) is within \(1 \%\) of its \(\begin{array}{ll}\text { (1) } & \text { (i) } 4 \text { is twithin } 1 \% \text { the } \\ \text { the the the the the the the the parition the parition }\end{array}\) limit, while \((1+5 / n)^{n}\) is still about \(12 \%\) from its limit. Give comparable accuracy estimates at \(x=1\) and \(a x=5\) for \(n=1000\).

Set $$a_{n}=\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\dots+\frac{n}{n^{2}}$$,Show that \(a_{n}\) is a Riemann sum for \(\int_{0}^{1} x d x .\) Docs the sequence \(a_{1}, a_{2}, \cdots\) converge? If so, to what?
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