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Chapter 10: Problem 133

For a sequence \(\left\\{a_{n}\right\\}\) the terms of even index are denoted by \(a_{2 k}\) and the terms of odd index by \(a_{2 k+1} .\) Prove that if \(a_{2 k} \rightarrow L\) and \(a_{2 k+1} \rightarrow L,\) then \(a_{n} \rightarrow L\).

Short Answer

Expert verified
If both subsequences converge to \( L \), then the entire sequence also converges to \( L \).

Step by step solution

01

Understanding the Problem

We have a sequence \( \{a_n\} \) where the even-indexed terms \( a_{2k} \) and the odd-indexed terms \( a_{2k+1} \) both converge to the limit \( L \). We need to show that the entire sequence also converges to \( L \).
02

Using Definition of Sequence Convergence

According to the definition of convergence, a sequence \( a_n \) converges to \( L \) if for every \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( n > N \), \( |a_n - L| < \epsilon \).
03

Convergence of Even-indexed Terms

By assumption, \( a_{2k} \rightarrow L \). This means that for every \( \epsilon > 0 \), there exists an integer \( N_1 \) such that for all even indexes \( 2k > N_1 \), \( |a_{2k} - L| < \epsilon \).
04

Convergence of Odd-indexed Terms

Similarly, since \( a_{2k+1} \rightarrow L \), for every \( \epsilon > 0 \), there exists an integer \( N_2 \) such that for all odd indexes \( 2k+1 > N_2 \), \( |a_{2k+1} - L| < \epsilon \).
05

Combining the Results

We need to find a single \( N \) such that for all \( n > N \), \(|a_n - L| < \epsilon \). Set \( N = \max(N_1, N_2) \). Then, for any \( n \), if \( n \) is even and \( n > N \), it follows that \( |a_{2k} - L| < \epsilon \). If \( n \) is odd and \( n > N \), then \(|a_{2k+1} - L| < \epsilon \). Thus, \( |a_n - L| < \epsilon \) for all \( n > N \).
06

Conclusion

Since for any \( \epsilon > 0 \), there exists an \( N \) such that for all \( n > N \), \( |a_n - L| < \epsilon \), we have shown that \( a_n \rightarrow L \). Thus, the entire sequence converges to the same limit \( L \).

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Key Concepts

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

Even-Indexed Terms
When we refer to even-indexed terms in a sequence, these are the terms like \(a_2, a_4, a_6, ...\). Mathematically, these can be represented as \(a_{2k}\), where \(k\) is an integer. To visualize this, think about picking every second element from an ordered list, starting from the second position.
  • If the sequence \(\{a_n\}\) represents someone's collection of marbles per month, the even-indexed terms are the number of marbles collected in February, April, June, and so on.
  • For a sequence to converge, it means that as \(k\) goes to infinity, all the even-indexed terms get closer and closer to a specific number, called the limit \(L\).
  • In our exercise, the even-indexed terms \(a_{2k}\) meet this condition as they converge to \(L\).
Understanding that even-indexed terms approach the same limit is crucial because it helps build the idea that the subsequence (a smaller part of the main sequence) behaves in a predictable manner. This property is often the first step in proving something about the entire sequence.
Odd-Indexed Terms
Odd-indexed terms in a sequence are those that fall at odd positions like \(a_1, a_3, a_5,...\). Representation-wise, they are denoted as \(a_{2k+1}\), where again \(k\) is an integer. These terms are similar to the even-indexed ones but start from the first item and skip every alternate term.
  • Imagine a series of streetlights turning on in a row, with odd-indexed lights lighting up first. The sequence could represent the brightness levels of those particular lights.
  • For convergence, as you consider more and more of these odd-indexed terms (as \(k\) increases), they too must approach the same limit \(L\).
  • In the given exercise, the condition \(a_{2k+1} \to L\) implies that this behavior is occurring.
This forms a critical component in the proof for sequence convergence because it shows that another subset of the sequence confirms the same behavior seen in the even-indexed terms. When both of these index types work towards the same end-point, this gives strength to the overall tendency of the sequence itself.
Limit of a Sequence
The concept of the limit is central to understanding sequence convergence. The goal is to show that the whole sequence \(\{a_n\}\) ends up surrounding a specific number \(L\) as you examine terms further in the sequence. This is what we call convergence.
  • In simpler terms, for any small number \(\epsilon > 0\), there is a point after which all the terms of the sequence are within \(\epsilon\) of \(L\).
  • The exercise showcases that when both the even-indexed and odd-indexed terms converge to \(L\), they combine to ensure that the whole sequence also converges.
  • So, if you ever find yourself thinking about how sequences stabilize or what their end result is as \(n\) increases without bound, you are pondering about their limit.
By considering both even and odd indexes converging, we can select an \(N\) that satisfies the convergence condition for all terms beyond some point. Thus, the limit \(L\) is effectively the target the entire sequence \(\{a_n\}\) aims for, proving the original sequence itself converges to \(L\).

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Most popular questions from this chapter

Assume that each sequence converges and find its limit. \(a_{1}=3, \quad a_{n+1}=12-\sqrt{a_{n}}\)

For approximately what values of \(x\) can you replace sin \(x\) by \(x-\left(x^{3} / 6\right)\) with an error of magnitude no greater than \(5 \times 10^{-4} ?\) Give reasons for your answer.

Use a CAS to perform the following steps for the sequences. 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 \(L ?\) \(a_{n}=\sin n\)

Use the definition of convergence to prove the given limit. \(\lim _{n \rightarrow \infty}\left(1-\frac{1}{n^{2}}\right)=1\)

Use the identity \(\sin ^{2} x=(1-\cos 2 x) / 2\) to obtain the Maclaurin series for \(\sin ^{2} x\). Then differentiate this series to obtain the Maclaurin series for \(2 \sin x \cos x .\) Check that this is the series for \(\sin 2 x\).
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