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Consider the formulas for the following sequences. Using a calculator, make a table with at least ten terms and determine a plausible value for the limit of the sequence or state that the sequence diverges. $$a_{n}=2 \tan ^{-1}(1000 n) ; n=1,2,3, \dots$$

Short Answer

Expert verified
If it converges, find the limit. Answer: The sequence \(a_n = 2 \tan^{-1}(1000 n)\) converges, and the limit as \(n\) approaches infinity is \(180^\circ\).

Step by step solution

01

Understanding the sequence and its notation

Here, we have a sequence \(a_n\) defined by the formula: $$a_{n}=2 \tan^{-1}(1000 n)$$ where \(n\) is a positive integer, and \(\tan^{-1}\) denotes the inverse tangent function (also denoted as \(\arctan\) or \(\operatorname{atan}\)).
02

Calculate the first ten terms of the sequence

To get a feel for the sequence, let's calculate its first ten terms using a calculator: 1. \(a_1 = 2 \tan^{-1}(1000 \cdot 1) \approx 179.45^\circ\) 2. \(a_2 = 2 \tan^{-1}(1000 \cdot 2) \approx 179.81^\circ\) 3. \(a_3 = 2 \tan^{-1}(1000 \cdot 3) \approx 179.91^\circ\) 4. \(a_4 = 2 \tan^{-1}(1000 \cdot 4) \approx 179.96^\circ\) 5. \(a_5 = 2 \tan^{-1}(1000 \cdot 5) \approx 179.98^\circ\) 6. \(a_6 = 2 \tan^{-1}(1000 \cdot 6) \approx 179.99^\circ\) 7. \(a_7 = 2 \tan^{-1}(1000 \cdot 7) \approx 179.99^\circ\) 8. \(a_8 = 2 \tan^{-1}(1000 \cdot 8) \approx 179.998^\circ\) 9. \(a_9 = 2 \tan^{-1}(1000 \cdot 9) \approx 179.998^\circ\) 10. \(a_{10} = 2 \tan^{-1}(1000 \cdot 10) \approx 179.998^\circ\)
03

Analyze the convergence/divergence of the sequence and estimate the limit if possible

Looking at the terms of the sequence, we can see that they are getting closer to \(180^\circ\) as \(n\) increases. This suggest that the sequence is likely to converge to \(180^\circ\). We can confirm this by analyzing the behavior of the \(\tan^{-1}\) function. Since \(\tan^{-1}(x)\) is an increasing function with a range of \((-90^\circ, 90^\circ)\), \(2 \tan^{-1}(x)\) will have a range of \((-180^\circ, 180^\circ)\). As \(x\) gets larger, \(2 \tan^{-1}(x)\) gets closer to \(180^\circ\). In our case, since \(1000n\) gets larger as \(n\) increases, we can conclude that the sequence converges and its limit is 180°: $$\lim_{n \to \infty} a_n = 180^\circ$$

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Key Concepts

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

Inverse Tangent Function
The inverse tangent function, commonly denoted as \(\tan^{-1}\), \(\arctan\), or \(\operatorname{atan}\), is a mathematical function that reverses the action of the tangent function. It's important to understand that for any given value \(x\), the \(\tan^{-1}(x)\) yields an angle \(\theta\) whose tangent is \(x\). This function is crucial in trigonometry and calculus as it surfaces in problems involving right triangles and in the integration of functions having a form of \(\frac{1}{1+x^2}\).

The inverse tangent is an odd function, meaning that \(\tan^{-1}(-x) = -\tan^{-1}(x)\) and is defined for all real numbers \(x\). The output of \(\arctan\) function is an angle typically given in radians within the range \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) or in degrees within the range \(\left(-90^\circ, 90^\circ\right)\). When computing the terms of our sequence \(a_n = 2 \tan^{-1}(1000 n)\), the \(\tan^{-1}\) function bounds each term to a maximum approaching \(90^\circ\), and by multiplying by 2, the upper boundary moves towards \(180^\circ\).

Understanding the nature of the inverse tangent function helps in predicting the behavior of sequences that utilize this function. As \(n\) grows larger, we observe that the values of \(2 \tan^{-1}(1000n)\) approach but never exceed \(180^\circ\), indicating a horizontal asymptote.
Convergence of Sequences
The convergence of sequences is a fundamental concept in mathematics, particularly in calculus. A sequence converges if the terms in the sequence approach a specific finite value as the sequence progresses to infinity. This specific finite value is called the limit of the sequence.

To determine convergence, one typically looks at the behavior of the sequence terms as \(n\) becomes large. If the terms get arbitrarily close to a number, the sequence is said to converge to that number. However, if the terms do not approach any particular value or go off to infinity, the sequence is said to diverge.

In our example with the sequence \(a_n = 2 \tan^{-1}(1000 n)\), we first calculate several terms. We notice these terms are getting closer to \(180^\circ\) as \(n\) increases. This pattern suggests convergence. In mathematical problems, looking at the sequence's trend through its initial terms helps formulate a hypothesis regarding its limit, which can then be confirmed through analytical methods or proofs.
Limit of a Sequence
Finally, let's discuss the limit of a sequence, which is intimately tied to the concept of convergence. The limit of a sequence is the value that the sequence's terms grow increasingly close to as \(n\) progresses towards infinity. Mathematically, for a sequence \(a_n\), if there exists a number \(L\) such that for every positive number \(\epsilon\), there is a corresponding \(N\) whereby for all \(n > N\), the terms \(a_n\) fall within \(\epsilon\) of \(L\), then the limit of \(a_n\) as \(n\) approaches infinity is \(L\), or \(\lim_{n \to \infty} a_n = L\).

For the sequence \(a_n\) given as an example, we compute enough terms to notice they inch towards \(180^\circ\) without ever actually reaching it. This behavior allows us to assert that \(\lim_{n \to \infty} a_n = 180^\circ\). It means as \(n\) gets larger and larger, the terms of the sequence get closer and closer to \(180^\circ\), but they theoretically will never precisely be \(180^\circ\), always falling just shy of it. While we estimated the limit by examining initial terms and using the properties of the inverse tangent function, in more complex cases, we may require rigorous mathematical proofs to establish the limit.

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Most popular questions from this chapter

James begins a savings plan in which he deposits \(\$ 100\) at the beginning of each month into an account that earns \(9 \%\) interest annually or, equivalently, \(0.75 \%\) per month. To be clear, on the first day of each month, the bank adds \(0.75 \%\) of the current balance as interest, and then James deposits \(\$ 100\). Let \(B_{n}\) be the balance in the account after the \(n\) th deposit, where \(B_{0}=\$ 0\). a. Write the first five terms of the sequence \(\left\\{B_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{B_{n}\right\\}\). c. How many months are needed to reach a balance of \(\$ 5000 ?\)

$$\text {Evaluate each series or state that it diverges.}$$ $$\sum_{k=1}^{\infty}\left(\sin ^{-1}(1 / k)-\sin ^{-1}(1 /(k+1))\right)$$

Consider the following sequences defined by a recurrence relation. Use a calculator, analytical methods, and/or graphing to make a conjecture about the limit of the sequence or state that the sequence diverges. $$a_{n+1}=\frac{1}{2} a_{n}+2 ; a_{0}=5$$

In the following exercises, two sequences are given, one of which initially has smaller values, but eventually "overtakes" the other sequence. Find the sequence with the larger growth rate and the value of \(n\) at which it overtakes the other sequence. $$a_{n}=\sqrt{n} \text { and } b_{n}=2 \ln n, n \geq 3$$

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b>1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n !>b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

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