Chapter 9: Problem 49
Prove that if \(\lim _{n \rightarrow \infty} a_{n}=0\) and \(\left\\{b_{n}\right\\}\) is bounded then \(\lim _{n \rightarrow \infty} a_{n} b_{n}=0\).
Short Answer
Expert verified
The product sequence \( a_n b_n \) approaches 0 as \( n \to \infty \) because \( a_n \) approaches 0 and \( b_n \) is bounded.
Step by step solution
01
Understanding the Limit of a Sequence
We start with the known limit of the sequence \( \{a_n\} \), which states that \( \lim_{n \to \infty} a_n = 0 \). This means for every \( \epsilon > 0 \), there exists an integer \( N \) such that for all \( n > N \), \( |a_n| < \epsilon \).
02
Defining a Bounded Sequence
The sequence \( \{b_n\} \) is given to be bounded. This implies there exists a positive constant \( M \) such that for all \( n \), \( |b_n| \leq M \).
03
Applying the Definition of Limit
To prove \( \lim_{n \to \infty} a_n b_n = 0 \), for every \( \epsilon > 0 \), we need to find an integer \( N \) such that for all \( n > N \), \( |a_n b_n| < \epsilon \).
04
Combining the Sequences
From the definition of limits and bounded sequences: since \( |a_n| < \epsilon/M \) for all \( n > N \) and \( |b_n| \leq M \), it follows that \( |a_n b_n| = |a_n| |b_n| < (\epsilon/M) \cdot M = \epsilon \).
05
Concluding the Proof
Hence, by choosing \( \epsilon/M \) suitably small, we ensure the product sequence \( \{a_n b_n\} \) satisfies the limit condition \( \lim_{n \to \infty} a_n b_n = 0 \), completing the proof.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bounded Sequences
When we say that a sequence \( \{b_n\} \) is bounded, it means that there is a specific limit to how large the absolute values of its terms can be. This can be compared to having a fence that the sequence cannot exceed. In formal terms, a sequence is bounded if there is a positive constant \( M \) such that for every term \( b_n \) in the sequence, \( |b_n| \leq M \).
This implies that no matter which term of the sequence you pick, its absolute value will always be less than or equal to this constant \( M \).
This implies that no matter which term of the sequence you pick, its absolute value will always be less than or equal to this constant \( M \).
- Bounded sequences do not necessarily converge; they only have an upper and a lower limit.
- This concept is useful in analysis since it sets a constraint we can rely on.
Convergence of Sequences
Understanding the convergence of sequences is central in mathematical analysis. A sequence \( \{a_n\} \) converges to a limit \( L \) if all of its terms get arbitrarily close to \( L \) as \( n \), the term number, becomes very large. Formally, \( \{a_n\} \) converges to \( L \) if for every small number \( \epsilon > 0 \), there is a sufficiently large integer \( N \) such that for all \( n > N \), the absolute difference \( |a_n - L| < \epsilon \).
- In our exercise, \( \lim_{n \to \infty} a_n = 0 \), meaning \( a_n \) gets closer and closer to 0 as \( n \) increases.
- Even small values for \( \epsilon \) can be chosen, but after a certain point \( N \), all terms satisfy \( |a_n| < \epsilon \).
Product of Sequences
The product of two sequences \( \{a_n\} \) and \( \{b_n\} \), denoted as \( \{a_n b_n\} \), consists of multiplying corresponding terms from each sequence. Understanding this concept is vital when one of the sequences is converging to zero.
If we have a sequence \( \{a_n\} \) that converges to zero and another sequence \( \{b_n\} \) that is bounded, the product sequence \( \{a_n b_n\} \) will also converge to zero. Here’s why:
If we have a sequence \( \{a_n\} \) that converges to zero and another sequence \( \{b_n\} \) that is bounded, the product sequence \( \{a_n b_n\} \) will also converge to zero. Here’s why:
- Since \( \{a_n\} \) converges to zero, for every small number \( \epsilon/M \) (derived by dividing \( \epsilon \) by the bound \( M \) of \( \{b_n\} \)), there is an integer \( N \) such that for all \( n > N \), \( |a_n| < \epsilon/M \).
- The sequence \( \{b_n\} \) being bounded means \( |b_n| \leq M \).
- Thus, \( |a_n b_n| = |a_n| |b_n| < (\epsilon/M) \cdot M = \epsilon \).