Problem 134 - Thomas Calculus

Question

Prove that a sequence \(\left\\{a_{n}\right\\}\) converges to 0 if and only if the sequence of absolute values \(\left\\{\left|a_{n}\right|\right\\}\) converges to \(0 .\)

Step-by-Step Solution

Answer

Both sequences converge to 0 simultaneously.

Step 1: Define Sequence Convergence

A sequence \( \{a_n\} \) is said to converge to a limit \( L \) if for every \( \varepsilon > 0 \), there exists a positive integer \( N \) such that for all \( n > N \), \( |a_n - L| < \varepsilon \). In this problem, we want \( a_n \to 0 \), so \( L = 0 \).

Step 2: Direct Implication

Assume \( \{a_n\} \) converges to 0. Then for every \( \varepsilon > 0 \), there exists \( N \) such that for all \( n > N \), \( |a_n - 0| < \varepsilon \). Therefore, this implies \( |a_n| < \varepsilon \), which means that the sequence \( \{|a_n|\} \) also converges to 0.

Step 3: Converse Implication

Assume \( \{|a_n|\} \) converges to 0. Then for every \( \varepsilon > 0 \), there exists \( N \) such that for all \( n > N \), \( ||a_n| - 0| < \varepsilon \), which simplifies to \( |a_n| < \varepsilon \). Thus, \( a_n \) converges to 0 since the absolute value of \( a_n \) is less than \( \varepsilon \).

Step 4: Conclude the Proof

Since we have shown both the direct implication (that \( a_n \to 0 \) implies \( |a_n| \to 0 \)) and the converse implication (that \( |a_n| \to 0 \) implies \( a_n \to 0 \)), it follows that the sequences \( \{a_n\} \) and \( \{|a_n|\} \) converge to 0 if and only if each other does.

Key Concepts

Absolute Value

The absolute value of a number tells us about its magnitude, regardless of its sign. It can be thought of as the number's distance from zero on a number line. Mathematically, the absolute value of a number, denoted as \( |a| \), simplifies any negative signs:

• If \( a \geq 0 \), then \( |a| = a \).
• If \( a < 0 \), then \( |a| = -a \).

The absolute value plays a significant role in sequences and convergence because it allows us to measure how close sequence elements are to 0, without worrying if they are positive or negative.

When we talk about sequences, like \( \{a_n\} \), considering their absolute values \( \{|a_n|\} \) assures us that the sequence is indeed getting closer to the limit, since absolute values cannot be negative. This is why in our original problem, understanding the convergence of the absolute value sequence \( \{|a_n|\} \) is crucial for addressing the convergence of the original sequence itself.

Direct Implication

Direct implication in mathematics is a way to establish that if one statement is true, another related statement must also be true. It's a form of logical reasoning that starts with a given premise and leads to a specific conclusion.

In the context of sequence convergence, direct implication helps us prove that if the sequence \( \{a_n\} \) converges to 0, then its absolute value sequence \( \{|a_n|\} \) also converges to 0. It works like this:

• Assume \( \{a_n\} \) converges to 0.
• This means that as \( n \) becomes very large, \( |a_n - 0| \), or \( |a_n| \), becomes smaller than any positive number \( \varepsilon \).
• Since \( |a_n| < \varepsilon \), \( \{|a_n|\} \) is shown to converge to 0 as well.

The direct implication is a straightforward approach, showing that if the original sequence heads towards 0, its magnitude does the same.

Converse Implication

Converse implication means showing the opposite of direct implication, i.e., proving that if the consequence is true, then the initial statement must also be true. It often helps in establishing a two-way relationship, known as "if and only if".

In our problem, the converse implication must prove that if \( \{|a_n|\} \) converges to 0, then \( \{a_n\} \) also converges to 0:

• Start with the assumption that \( \{|a_n|\} \) converges to 0. This means for every \( \varepsilon > 0 \), there exists an \( N \) where all sequence elements beyond \( N \) are smaller in absolute value than \( \varepsilon \).
• Since \( |a_n| < \varepsilon \), the actual terms of the sequence, \( a_n \), must also stay within \( \varepsilon \) after a certain point.
• This proves that \( a_n \), without considering absolute values, converges to 0.

By establishing this converse, we effectively link the convergence of \( \{a_n\} \) and \( \{|a_n|\} \) as equivalent, rounding off the proof logically and soundly.