/*! 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 52 Which of the sequences \(\left\\... [FREE SOLUTION] | 91Ó°ÊÓ

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Which of the sequences \(\left\\{a_{n}\right\\}\) in Exercises \(23-84\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\sqrt[n]{n^{2}} $$

Short Answer

Expert verified
The sequence converges to 1.

Step by step solution

01

Review Sequence Definition

The given sequence is \( a_n = \sqrt[n]{n^2} \). We need to determine whether this sequence converges or diverges, and if it converges, find its limit.
02

Apply Limit Property

Examine the term \( a_n = n^{2/n} \). This can be rewritten as \( n^{2/n} = e^{\ln(n^{2/n})} = e^{\frac{2}{n} \ln n} \). To find the limit, evaluate \( \lim_{n \to \infty} \frac{2 \ln n}{n} \).
03

Evaluate the Limit of the Exponent

Use L'Hôpital's Rule to evaluate \( \lim_{n \to \infty} \frac{2 \ln n}{n} \) because it's in the form \( \frac{\infty}{\infty} \). Differentiating the numerator and denominator gives \( \lim_{n \to \infty} \frac{2 \cdot (1/n)}{1} = \lim_{n \to \infty} \frac{2}{n} = 0 \).
04

Conclude on Sequence Convergence

Since we found that \( \lim_{n \to \infty} \frac{2 \ln n}{n} = 0 \), it follows that \( \lim_{n \to \infty} e^{\frac{2}{n} \ln n} = e^0 = 1 \). Hence, the sequence \( a_n = \sqrt[n]{n^2} \) converges to 1.

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Key Concepts

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

Limits
Limits are a fundamental concept in calculus. They help us understand the behavior of functions or sequences as they approach a certain point or infinity. When we say that a sequence converges to a limit, we mean that the sequence approaches a particular value as its index (often denoted by \(n\)) becomes very large.

For sequences, evaluating limits generally involves finding the behavior of the sequence \(a_n\) as \(n\) goes to infinity. We analyze whether \(a_n\) begins to approach and stabilize around a particular number. If it does, the sequence is said to converge to that number, which is its limit.

In our example, the sequence was \(a_n = \sqrt[n]{n^2}\). We had to find if this sequence had a limit as \(n\) becomes very large. By examining the limit, we determined it converges to 1. This means, as \(n\) grows infinitely large, \(a_n\) gets closer and closer to 1, and stays there.
L'Hôpital's Rule
L'Hôpital's Rule is a useful tool in calculus for evaluating limits that initially seem indeterminate. Indeterminate forms often come in the guise of \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

L'Hôpital's Rule states that if you have an indeterminate form, \(\lim_{x \to c} \frac{f(x)}{g(x)}\), you can differentiate the numerator and the denominator separately and then take the limit again: \(\lim_{x \to c} \frac{f'(x)}{g'(x)}\), provided the new limit exists.

In the solution, we used L'Hôpital’s Rule to tackle the exponent \(\lim_{n \to \infty} \frac{2 \ln n}{n}\), which was initially in the form \(\frac{\infty}{\infty}\). By differentiating both the numerator \(2 \ln n\) and the denominator \(n\), we successfully found that the limit is 0. This outcome helped us conclude that the overall sequence limit is 1.
Exponential Functions
Exponential functions are powerful mathematical tools frequently seen in scientific and financial models, among others. In essence, an exponential function has the form \(e^x\), where \(e\) is the base of the natural logarithms, approximately equal to 2.718.When dealing with sequences or limits, exponential functions often facilitate the simplification and analysis of complex expressions. In calculus, converting expressions into exponential forms can reveal more about their behavior as the variable grows.

For the sequence \(a_n = \sqrt[n]{n^2}\), rewriting \(n^{2/n}\) as an exponential helped us handle the sequence's limit more dynamically. We transformed it into \(e^{\frac{2}{n} \ln n}\). Then, by solving for the limit of its exponent, we figured out that the sequence converges. This underscores the utility of exponential functions in mathematical analysis. In our case, this transformation indicated that \(a_n\) stabilizes at a limit of 1 as \(n\) becomes very large.

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Most popular questions from this chapter

Estimating Pi About how many terms of the Taylor series for \(\tan ^{-1} x\) would you have to use to evaluate each term on the right-hand side of the equation $$ \pi=48 \tan ^{-1} \frac{1}{18}+32 \tan ^{-1} \frac{1}{57}-20 \tan ^{-1} \frac{1}{239} $$ with an error of magnitude less than \(10^{-6} ?\) In contrast, the convergence of \(\sum_{n=1}^{\infty}\left(1 / n^{2}\right)\) to \(\pi^{2} / 6\) is so slow that even 50 terms will not yield two-place accuracy.

Find series solutions for the initial value problems in Exercises \(15-32\) . $$ y^{\prime \prime}+y=0, \quad y^{\prime}(0)=0 \text { and } y(0)=1 $$

Show that if \(\Sigma_{n=1}^{\infty} a_{n}\) converges absolutely, then $$\left|\sum_{n=1}^{\infty} a_{n}\right| \leq \sum_{n=1}^{\infty}\left|a_{n}\right|$$

In Exercises \(45-48\) , estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{(0.01)^{n}}{n} $$

Use a CAS to perform the following steps for the sequences in Exercises \(129-140 .\) a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\frac{n^{41}}{19^{n}} $$

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