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Chapter 10: Problem 84

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

Short Answer

Expert verified
The sequence converges to 1 as \( n \to \infty \).

Step by step solution

01

Write the sequence formula

The sequence you are given is \( a_n = \sqrt[n]{n^2 + n} \). This expression represents the \( n \)-th term of the sequence \( \{a_n\} \).
02

Simplify the expression inside the root

Notice that as \( n \) becomes very large, the term \( n^2 + n \) can be approximated by \( n^2 \) since \( n^2 + n \equiv n^2(1 + \frac{1}{n}) \). This simplification is helpful to explore the behavior of the sequence as \( n \to \infty \).
03

Rewrite the sequence using the simplification

Given the simplification, you can rewrite the expression as \( a_n = \sqrt[n]{n^2(1 + \frac{1}{n})} \). By property of roots, this can be expressed as \( a_n = \left(n^2\right)^{1/n}(1+\frac{1}{n})^{1/n} \).
04

Further simplify \((n^2)^{1/n}\)

Recall that \((n^2)^{1/n} = n^{2/n} = (e^{\ln(n^2)})^{1/n} = e^{\frac{2\ln(n)}{n}}\). As \( n \to \infty \), \( \frac{2\ln(n)}{n} \to 0\). Therefore, \( n^{2/n} \to e^0 = 1 \).
05

Simplify \((1+\frac{1}{n})^{1/n}\)

For the expression \( (1+\frac{1}{n})^{1/n} \), we can use the fact that \( (1+x)^{x} \to 1 \) as \( x \to 0 \). Thus, as \( n \to \infty \), \( (1+\frac{1}{n})^{1/n} \to 1 \).
06

Determine the limit of the sequence

Combining results from steps 4 and 5, we get \( a_n = n^{2/n}(1+\frac{1}{n})^{1/n} \to 1 \times 1 = 1 \). This confirms that the sequence converges to 1.

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Key Concepts

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

Limit of a Sequence
Understanding the limit of a sequence is crucial in determining whether it converges or not. The limit of a sequence \(\) is defined as the value that the terms of the sequence approach as \( n\) becomes very large. In the exercise, the sequence is given by \(a_n = \sqrt[n]{n^2 + n}\). To find its limit, we analyze the behavior of the sequence as \( n\) approaches infinity.
  • If a sequence approaches a single finite value, it converges to that limit.
  • If it doesn't settle on any value, it diverges.
For the sequence \(a_n\), through simplification and evaluation, we find that it converges to 1. This means, as \( n\) gets infinitely large, the term \(a_n\) gets closer and closer to 1.
Sequence Simplification
Sequence simplification is a technique to understand the long-term behavior of a sequence. In essence, it's about making complicated sequences easier to analyze. Simplifying the sequence \( a_n = \sqrt[n]{n^2 + n}\) involves approximating the inner expression \( n^2 + n\) when \( n\) is very large.
  • First, observe that \( n^2 + n\) can be simplified to \( n^2(1 + \frac{1}{n})\). Here, \( n^2\) dominates since it's much larger than \( n\) when \( n\) is large.
  • Next, use properties of roots: \( a_n = ((n^2)^{1/n})(1 + \frac{1}{n})^{1/n}\).
This simplification is essential as it allows us to find the sequence's limit more straightforwardly by focusing on the dominant term in the sequence.
n-th Term Behavior
Analyzing the \( n \)-th term behavior of a sequence helps determine how each term changes as \( n\) increases. For the sequence \( a_n = \sqrt[n]{n^2 + n}\), let's focus on the components:
  • The term \( (n^2)^{1/n}\) can be rewritten as \( n^{2/n} = e^{\frac{2\ln(n)}{n}}\), which approaches 1 since \( \frac{2\ln(n)}{n}\) tends to 0.
  • Similarly, \( (1 + \frac{1}{n})^{1/n}\) goes to 1 as it uses the approximation \( (1+x)^x \to 1\) as \( x \to 0\).
Combining these results shows that each term \( a_n\) tends to 1, confirming the overall behavior as \( n\) grows.
Infinite Sequences
Infinite sequences are sequences that continue indefinitely as \( n\) progresses to infinity. They are a foundation in the study of calculus and analysis. Convergence or divergence in these sequences tells us about the potential for finding a limit.
  • In our sequence \( a_n = \sqrt[n]{n^2 + n}\), viewing it as an infinite sequence allows us to examine how its terms behave over the entirety of their lifespan.
  • By determining its limit, we conclude that this sequence doesn't just wander aimlessly but homes in on a specific number 鈥� in this case, 1.
Understanding infinite sequences can thus help us make predictions about long-term trends and behaviors of mathematical functions.

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Most popular questions from this chapter

Assume that each sequence converges and find its limit. \(a_{1}=2, \quad a_{n+1}=\frac{72}{1+a_{n}}\)

If \(\Sigma a_{n}\) converges and \(a_{n} > 0\) for all \(n\), can anything be said about \(\Sigma\left(1 / a_{n}\right) ?\) Give reasons for your answer.

Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials \(P_{1}(x), P_{2}(x),\) and \(P_{3}(x)\) at \(\bar{x}=0\) Step 3: Calculate the ( \(n+1\) )st derivative \(f^{(n+1)}(c)\) associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of \(c\) over the specified interval and estimate its maximum absolute value, \(M\) Step 4: Calculate the remainder \(R_{n}(x)\) for each polynomial. Using the estimate \(M\) from Step 3 in place of \(f^{(n+1)}(c),\) plot \(R_{n}(x)\) over the specified interval. Then estimate the values of \(x\) that answer question (a). Step 5: Compare your estimated error with the actual error \(E_{n}(x)=\left|f(x)-P_{n}(x)\right|\) by plotting \(E_{n}(x)\) over the specified interval. This will help answer question (b). Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. $$f(x)=(\cos x)(\sin 2 x), \quad|x| \leq 2$$

a. Show that if two power series \(\Sigma_{n=0}^{\infty} a_{n} x^{n}\) and \(\Sigma_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n\) \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(a_{n}\) and \(b_{n}\) both equal \(f^{(n)}(0) /(n !)\) b. Show that if \(\Sigma_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n\)

Use the definition of \(e^{i \theta}\) to show that for any real numbers \(\theta, \theta_{1}\) and \(\theta_{2}\) a. \(\quad e^{i \theta_{1}} e^{i \theta_{2}}=e^{i\left(\theta_{1}+\theta_{2}\right)}\) b. \(e^{-i \theta}=1 / e^{i \theta}\)
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