/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Use a graphing utility or CAS to... [FREE SOLUTION] | 91Ó°ÊÓ

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Use a graphing utility or CAS to plot the first 15 terms of the sequence. Determine whether the sequence converges, and if it does, give the limit. (a) \(\frac{n}{\sqrt[n]{n !}}\) (b) \(n^{2} \sin (\pi / n)\)

Short Answer

Expert verified
Sequence (a) converges to the limit \(e\), while sequence (b) diverges as the limit goes to infinity.

Step by step solution

01

Define the sequence for (a)

For the sequence (a), the given formula is: \(a_{n} = \frac{n}{\sqrt[n]{n !}}\)
02

Plot the first 15 terms for (a)

Using a graphing utility or CAS, plot the first 15 values of the sequence (a) to visualize its behavior.
03

Analyze the sequence graph for (a)

Observe the plotted sequence to see if the terms seem to be getting closer together or further apart. Notice any patterns in the behavior of the sequence as the number of terms increases.
04

Determine the limit for (a)

To determine if the sequence converges, we need to find the limit as \(n\) approaches infinity. Stirling's approximation can be used to approximate the limit, which states: \(n! \approx \sqrt{2 \pi n} * \left(\frac{n}{e}\right)^n\) Taking this approximation into account, the limit of the sequence becomes: \(\lim_{n \to \infty} \frac{n}{\sqrt[n]{n !}} \approx \lim_{n \to \infty} \frac{n}{\sqrt[n]{\sqrt{2 \pi n} * \left(\frac{n}{e}\right)^n}}\) Simplify the limit expression and calculate the limit, which will give the value \(e\).
05

Verify the convergence for (a)

Since the limit of the sequence exists and equals \(e\), the sequence converges. The graph of the plotted terms should support this conclusion visually. ##Sequence (b)##
06

Define the sequence for (b)

For the sequence (b), the given formula is: \(b_{n} = n^{2} \sin (\pi / n)\)
07

Plot the first 15 terms for (b)

Using a graphing utility or CAS, plot the first 15 values of the sequence (b) to visualize its behavior.
08

Analyze the sequence graph for (b)

Observe the plotted sequence to see if the terms seem to be getting closer together or further apart. Notice any patterns in the behavior of the sequence as the number of terms increases.
09

Determine the limit for (b)

To determine if the sequence converges, we need to find the limit as \(n\) approaches infinity. \(\lim_{n \to \infty} n^{2} \sin (\pi / n)\) Apply the squeeze theorem: since \(0 \leq \sin (\pi / n) \leq 1\), we have \(0 \leq n^{2} \sin (\pi / n) \leq n^{2}\). As \(n \to \infty\), \(n^{2}\) also approaches infinity, so the limit of the sequence is infinity.
10

Verify the convergence for (b)

Since the limit of the sequence goes to infinity, the sequence diverges. The graph of the plotted terms should also support this conclusion visually.

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Key Concepts

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

Understanding Sequence Limits
When studying sequences in mathematics, one key concept that emerges is the 'limit of a sequence'. This refers to the value that the sequence's terms get closer to as the term number increases indefinitely. For example, consider the sequence defined by the formula \( a_n = \frac{n}{\sqrt[n]{n!}} \). To understand if this sequence converges, we look for the limit as \( n \) approaches infinity.

Convergence means that there is a specific number that the terms of the sequence get infinitely close to. If such a number exists, the sequence is said to converge to that number, also known as the 'limit' of the sequence. The graphical plot can offer insights into the behavior of the sequence and whether terms are 'squeezing' together towards a single value or spreading apart.

For sequence (a), we deduce convergence by showing that as \( n \) increases, the terms of the sequence approach a particular value. This alignment with a definite value is pivotal in understanding the fundamental nature of sequences in calculus.
Exploring Stirling's Approximation
In higher mathematics, 'Stirling's approximation' is an essential tool for estimating the factorial of large numbers. It is especially useful when factorials are part of complex functions or sequences. The formula states that \( n! \) can be approximated by \( \sqrt{2 \pi n} * \left(\frac{n}{e}\right)^n \), where \( e \) is the base of the natural logarithm.

This approximation becomes incredibly accurate as \( n \) grows large. In our sequence (a), Stirling's approximation is utilized to find the limit of \( \frac{n}{\sqrt[n]{n!}} \) as \( n \) approaches infinity. Through simplifying the expression using Stirling's approximation, we can then calculate the limit, which in this case, yields the value \( e \). This technique is a brilliant example of how a complex concept can be made simpler through a powerful approximation and depicts the synergy between different areas of mathematics.
Applying the Squeeze Theorem
The 'squeeze theorem' (also known as the 'sandwich theorem') is an invaluable concept in calculus for determining the limits of sequences, especially when the sequence itself is difficult to evaluate directly. The theorem states that if a sequence is 'squeezed' between two other sequences whose limits are the same, then it must also converge to the same limit.

In relation to sequence (b) from the exercise, \( b_n = n^2 \sin(\pi / n) \), we apply the squeeze theorem by showing that \( 0 \leq \sin(\pi / n) \leq 1 \) and consequently \( 0 \leq n^2 \sin(\pi / n) \leq n^2 \). As \( n \) tends to infinity, the term \( n^2 \) increases without bound. Therefore, the limit of the sequence is infinity, confirming divergence.

This theorem is an incredible demonstration of how boundaries can be used to establish limits and facilitates the understanding of sequences' behavior at infinity.

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Most popular questions from this chapter

(a) Let \(a_{1}, a_{2}, \cdots\) be a convergent sequence. Prove that $$\lim _{n \rightarrow \infty}\left(a_{n}-a_{n-1}\right)=0$$. (b) W'ilat can you say about the converse? That is, suppose that \(a_{1}, a_{2}, \cdots\) is a sequence for which $$\lim _{n \rightarrow \infty}\left(a_{n}-a_{n-1}\right)=0$$. Does \(a_{1} \cdot a_{2}, \cdots\) necessarily converge? If so, prove it; jf not, give a counterexample.

Give the first six terms of the sequence and then give the \(n\) th term. $$a_{1}=1 ; \quad a_{n+1}=2 a_{n}+1$$.

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 \(S\).

(a) For what values of \(r\) is $$\int_{0}^{\infty} x^{r} e d x$$ convergent? (b) Show by induction that $$\int_{0}^{\infty} x^{n} e^{-x} d x=n !, n=1,2,3, \cdots$$

Sketch the curve, specifying all vertical and horizontal asymptotes. $$y=x e^{x}$$

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