Problem 41
Question
Assuming that \(u_{1}=0\) and \(u_{n+1}=1.1^{u_{n}}\) determine a convergent sequence, find \(\lim _{n \rightarrow \infty} u_{n}\) to four decimal places.
Step-by-Step Solution
Verified Answer
The sequence converges to 1.0000.
1Step 1: Understand the Sequence Definition
We need to determine if the sequence \(u_n\), defined by \(u_{1}=0\) and \(u_{n+1}=1.1^{u_{n}}\), converges and find its limit. Each term in the sequence is generated from the previous term by substituting it into the function \(1.1^{x}\).
2Step 2: Check for Monotonicity and Boundedness
To determine convergence, analyze the behavior of the function \(f(u) = 1.1^u\). As \(u_n = 1.1^{u_{n-1}}\), we can see that if \(u_{n} < 1\), then \(u_{n+1} = 1.1^{u_n} > u_n\), so it increases. If \(u_n > 1\), then \(u_{n+1} < u_n\), so it decreases. Therefore, the sequence oscillates around the steady point \(u_* = 1\). Since each iteration refines the estimate of the limit, the sequence is bounded by 0 and 1.1 and will converge.
3Step 3: Determine the Limit
Assume the sequence converges to a limit \(L\). Therefore, \[L = 1.1^L\]. Solve \(x = 1.1^x\) by looking for solutions graphically or using an iterative approach. The iterative approach finds that the sequence converges to \(L = 1\), verifying it satisfies the equation \(1 = 1.1^1\).
4Step 4: Verify Decimal Places
Check the convergence to four decimal places using trial terms. Calculate successive terms until the change in terms is less than 0.0001. The calculation confirms that the sequence converges to a limit at which \(u_n\) stabilizes around 1.
Key Concepts
MonotonicityBoundednessIterative Methods
Monotonicity
Monotonicity refers to a property where a sequence either never increases or never decreases as it progresses. Understanding whether a sequence is monotonic is vital in analyzing its convergence behavior. For the given sequence, where each term is calculated by raising 1.1 to the power of the previous term, monotonicity can initially be a bit complex to unravel.
However, we have a specific condition - if the sequence term \(u_n\) is less than 1, then the following term \(u_{n+1} = 1.1^{u_n}\) will be greater than \(u_n\). This means the sequence is increasing in such cases. Conversely, if \(u_n > 1\), the next term \(u_{n+1}\) becomes smaller, indicating a decreasing pattern.
This alternating behavior indicates that the sequence is neither strictly increasing nor decreasing but rather oscillates when approaching a stable value. Recognizing this oscillation is key to understanding that such behavior assists in the sequence's convergence, helping to steady around a limit.
However, we have a specific condition - if the sequence term \(u_n\) is less than 1, then the following term \(u_{n+1} = 1.1^{u_n}\) will be greater than \(u_n\). This means the sequence is increasing in such cases. Conversely, if \(u_n > 1\), the next term \(u_{n+1}\) becomes smaller, indicating a decreasing pattern.
This alternating behavior indicates that the sequence is neither strictly increasing nor decreasing but rather oscillates when approaching a stable value. Recognizing this oscillation is key to understanding that such behavior assists in the sequence's convergence, helping to steady around a limit.
Boundedness
A sequence is bounded if there are real numbers that act as limits within which all terms of the sequence must lie. Boundedness is a significant concept in evaluating the convergence of sequences; it ensures that the sequence doesn't diverge to infinity.
For the sequence defined in the exercise, it begins at 0 and each subsequent term is calculated through the function \(1.1^{x}\), which indicates rapid diminishment. As indicated at step 2 of the solution, the sequence shows it remains constrained between 0 and 1.1. This implies it's bounded.
The critical insight here is that once the sequence illustrates behavior of consistent bounding around a point, such as oscillating between zero and a little over one, it's a strong indicator that the sequence will indeed converge. Establishing these bounds simplifies the problem of predicting long-term behavior as all terms of the sequence are known to not exceed these specific bounds.
For the sequence defined in the exercise, it begins at 0 and each subsequent term is calculated through the function \(1.1^{x}\), which indicates rapid diminishment. As indicated at step 2 of the solution, the sequence shows it remains constrained between 0 and 1.1. This implies it's bounded.
The critical insight here is that once the sequence illustrates behavior of consistent bounding around a point, such as oscillating between zero and a little over one, it's a strong indicator that the sequence will indeed converge. Establishing these bounds simplifies the problem of predicting long-term behavior as all terms of the sequence are known to not exceed these specific bounds.
Iterative Methods
Iterative methods are commonly used for finding approximate solutions to otherwise complex mathematical problems like calculating limits of sequences. It involves taking initial guesses and refining them through a process of iteration until precision is sufficiently achieved.
In this exercise, the sequence \(u_{n+1} = 1.1^{u_n}\) serves as its own iterative step. Each application of the function \(1.1^{x}\) refines the estimate of the limit by getting a new term closer to the anticipated boundary. Iterative methods often involve calculations done multiple times, using the output of one operation as the input to the next.
By computing terms recursively, we can determine when the terms no longer significantly change by noting when the difference between consecutive terms drops below a chosen small number (0.0001 here). This confirms convergence and determines that \(L = 1\) is the limit to which it converges in four decimal precision.
In this exercise, the sequence \(u_{n+1} = 1.1^{u_n}\) serves as its own iterative step. Each application of the function \(1.1^{x}\) refines the estimate of the limit by getting a new term closer to the anticipated boundary. Iterative methods often involve calculations done multiple times, using the output of one operation as the input to the next.
By computing terms recursively, we can determine when the terms no longer significantly change by noting when the difference between consecutive terms drops below a chosen small number (0.0001 here). This confirms convergence and determines that \(L = 1\) is the limit to which it converges in four decimal precision.
Other exercises in this chapter
Problem 41
Prove that if \(\sum_{k=1}^{\infty} a_{k}^{2}\) and \(\sum_{k=1}^{\infty} b_{k}^{2}\) both converge then \(\sum_{k=1}^{\infty} a_{k} b_{k}\) converges absolutel
View solution Problem 41
$$ \text { Prove: If } \sum_{k=1}^{\infty} a_{k} \text { diverges, so does } \sum_{k=1}^{\infty} c a_{k} \text { for } c \neq 0 \text { . } $$
View solution Problem 42
Test for convergence or divergence using the Root Test. (a) \(\sum_{n=2}^{\infty}\left(\frac{1}{\ln n}\right)^{n}\) (b) \(\sum_{n=1}^{\infty}\left(\frac{n}{3 n+
View solution Problem 43
Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems \(43-48\) to see that you get the same answers
View solution