Problem 89 - Calculus: Early Transcendentals

Question

Prove that if \( \lim_{n \to \infty} a_n = 0 \) and \( \left \\{ b_n \right \\} \) is bounded, then \( \lim_{n \to\infty} (a_n b_n) = 0. \)

Step-by-Step Solution

Answer

The limit of \( a_n b_n \) is 0 as \( n \to \infty \).

Step 1: Understanding the Problem

We are given that the sequence \(a_n\) converges to 0 and the sequence \(b_n\) is bounded. We need to prove that the product of these sequences, \(a_n b_n\), also converges to 0 as \(n\) approaches infinity.

Step 2: Definition of Convergence

Recall that \( \lim_{n \to \, \infty} a_n = 0 \) means for every \( \varepsilon > 0 \), there exists a positive integer \( N_1 \) such that for all \( n > N_1 \), \( |a_n| < \varepsilon \).

Step 3: Definition of Bounded Sequence

A sequence \( \{ b_n \} \) is bounded if there exists some constant \( M > 0 \) such that for all \( n \), \( |b_n| \leq M \).

Step 4: Multiply and Simplify

Consider the product \( |a_n b_n| = |a_n| |b_n| \). Since \( |a_n| < \varepsilon \) for all \( n > N_1 \) and \( |b_n| \leq M \), we have \( |a_n b_n| < \varepsilon M \).

Step 5: Choose a New Epsilon

To ensure \( |a_n b_n| < \varepsilon \), let us choose \( \varepsilon' = \frac{\varepsilon}{M} \). Since \( |a_n| < \varepsilon' \) eventually, it follows that \( |a_n b_n| < \varepsilon \).

Step 6: Conclusion of the Proof

For every \( \varepsilon > 0 \), there exists \( N_1 \) such that for all \( n > N_1 \), \(|a_n b_n| < \varepsilon\). Therefore, by the definition of convergence, \( \lim_{n \to \infty} (a_n b_n) = 0 \).

Key Concepts

Sequence Convergence

Sequence convergence is an essential concept in calculus and analysis. When we say a sequence \(a_n\) converges to a limit, \(L\), it means the terms of the sequence get indefinitely close to \(L\) as \(n\) goes to infinity. Mathematically, a sequence \(a_n\) is said to converge to \(L\) if for every \(\varepsilon > 0\), there exists a natural number \(N\) such that \(|a_n - L| < \varepsilon \) for all \(n > N\). In simpler terms:

• The sequence closely approaches \(L\) as \(n\) gets larger and larger.
• Given any small positive distance \(\varepsilon\), eventually all terms of the sequence will lie within that distance from \(L\).

This means convergence is about how close the sequence can get to the limit and not necessarily at which term it hits the limit. This understanding is crucial for working with the concept of limits and continuity in mathematics.

Bounded Sequence

A bounded sequence is one where all its terms are confined within some fixed range. Specifically, a sequence \(\{b_n\}\) is bounded if there is a real number \(M > 0\) such that \(|b_n| \leq M\) for all \(n\). This means:

• There exists an upper bound on the magnitudes of all terms of the sequence.
• No term in the sequence can exceed this bound in absolute value.

Boundedness ensures that though the terms may oscillate or change, they do not diverge to infinity or fall below a specific negative value. It provides constraints that help in various mathematical calculations and theorems. In our exercise, \(b_n\) being bounded plays a pivotal role in helping determine the behavior of the product of sequences.

Product of Sequences

The product of sequences involves multiplying each corresponding term of two sequences together to form a new sequence. If we have two sequences, \(a_n\) and \(b_n\), their product sequence is given by \(a_n b_n\). Understanding how the product behaves depends on the individual properties of each sequence:

• If \(a_n\) converges to zero while \(b_n\) is bounded, the product sequence \(a_n b_n\) will converge to zero.
• This is because the \(a_n\) terms will shrink towards zero, while the bound on \(b_n\) ensures no term can grow unboundedly.

Thus, the convergence of \(a_n b_n\) is guaranteed under these conditions. This property is particularly useful in mathematical analysis, helping to simplify complex expressions and calculations involving products of terms. The reasons involve the nature of convergence and how constraints (being bounded) limit the growth of sequences.