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Chapter 10: Problem 83

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. \(a_{n}=\left(\frac{1}{3}\right)^{n}+\frac{1}{\sqrt{2^{n}}}\)

Short Answer

Expert verified
The sequence converges to 0.

Step by step solution

01

Analyze components of the sequence

The given sequence is \(a_n = \left(\frac{1}{3}\right)^n + \frac{1}{\sqrt{2^n}}\). It consists of two parts: \(\left(\frac{1}{3}\right)^n\) and \(\frac{1}{\sqrt{2^n}}\). We need to analyze the convergence of each component separately.
02

Examine \(\left(\frac{1}{3}\right)^n\)

The term \(\left(\frac{1}{3}\right)^n\) is a geometric sequence with a common ratio of \(\frac{1}{3}\), where \(0 < \frac{1}{3} < 1\). Geometric sequences with ratios between 0 and 1 converge to 0 as \(n\) approaches infinity.
03

Examine \(\frac{1}{\sqrt{2^n}}\)

The term \(\frac{1}{\sqrt{2^n}}\) can be rewritten as \((2^{-1/2})^n\). This is another geometric sequence with a common ratio of \(2^{-1/2}\), where \(0 < 2^{-1/2} < 1\). Similarly, this sequence also converges to 0 as \(n\) approaches infinity.
04

Combine results to evaluate \(a_n\)

Since both components \(\left(\frac{1}{3}\right)^n\) and \(\frac{1}{\sqrt{2^n}}\) converge to 0, their sum also converges to 0. Therefore, the sequence \(a_n\) converges.
05

Find the limit of the convergent sequence

Given that each term in the sequence \(a_n = \left(\frac{1}{3}\right)^n + \frac{1}{\sqrt{2^n}}\) tends to 0, the limit of the sequence as \(n\) approaches infinity is 0.

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Key Concepts

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

Geometric Sequence
A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, consider the sequence \(a_n = \left(\frac{1}{3}\right)^n\). Here, each term is obtained by multiplying the previous term by \(\frac{1}{3}\). Since this ratio is between 0 and 1, the sequence converges towards 0.
  • In a geometric sequence, if the common ratio \(r\) satisfies \(0 < r < 1\), the sequence converges to 0.
  • If \(r = 1\), the sequence is constant and converges to the exact value of its terms.
  • If \(r > 1\) or \(r < -1\), the sequence diverges, meaning it grows to infinity or oscillates without settling at a particular value.
Understanding the behavior of geometric sequences is crucial in calculus and helps in determining the limits of such sequences.
Limits in Calculus
Limits in calculus describe the behavior of a function or sequence as its input approaches a particular point. For sequences, we often look at the behavior as the number of terms (\(n\)) approaches infinity. We write this as \( \lim_{n \to \infty} \).
  • Applying the concept of limits to the sequence \(a_n = \left(\frac{1}{3}\right)^n + \frac{1}{\sqrt{2^n}}\), we see both components have limits of 0 as \(n\) approaches infinity.
  • This concludes that \( \lim_{n \to \infty} a_n = 0\).
This specific application of limits is what helps mathematically prove that our sequence converges. It also illustrates how limits allow for the precise handling of behaviors of functions and sequences, validating intuition through rigorous mathematical grounding.
Convergent and Divergent Sequences
In mathematical analysis, sequences can either converge or diverge depending on their behavior as \(n\) tends to infinity.
  • A sequence converges if it approaches a finite limit or specific point as \(n\) becomes very large.
  • Alternatively, a sequence diverges if it moves away without approaching any fixed limit.
For example, our sequence \(a_n = \left(\frac{1}{3}\right)^n + \frac{1}{\sqrt{2^n}}\) showcases both components converging to 0. Thus, the sequence as a whole converges to 0. Recognizing whether a sequence converges or diverges is essential for evaluating long-term trends
and behaviors in mathematical analysis and has a wide application in various fields such as computational mathematics and physics.

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Most popular questions from this chapter

Integrate the binomial series for \(\left(1-x^{2}\right)^{-1 / 2}\) to show that for \(|x|<1\) $$ \sin ^{-1} x=x+\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2 n)} \frac{x^{2 n+1}}{2 n+1} $$

Which of the sequences converge, and which diverge? Give reasons for your answers. \(a_{n}=n-\frac{1}{n}\)

The (second) second derivative test Use the equation $$f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}\left(c_{2}\right)}{2}(x-a)^{2}.$$to establish the following test. Let \(f\) have continuous first and second derivatives and suppose that \(f^{\prime}(a)=0 .\) Then a. \(f\) has a local maximum at \(a\) if \(f^{\prime \prime} \leq 0\) throughout an interval whose interior contains \(a\) b. \(f\) has a local minimum at \(a\) if \(f^{\prime \prime} \geq 0\) throughout an interval whose interior contains \(a\).

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

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. $$f(x)=(\cos x)(\sin 2 x), \quad|x| \leq 2$$
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