Problem 64 - Thomas Calculus

Question

Show that if \(\sum_{n=1}^{\infty} a_{n}\) converges absolutely, then $$\left|\sum_{n=1}^{\infty} a_{n}\right| \leq \sum_{n=1}^{\infty}\left|a_{n}\right|$$

Step-by-Step Solution

Answer

The triangle inequality and taking limits show \( \left|\sum_{n=1}^{\infty} a_n\right| \leq \sum_{n=1}^{\infty}|a_n| \).

Step 1: Understand Absolute Convergence

A series \( \sum_{n=1}^{\infty} a_n \) converges absolutely if \( \sum_{n=1}^{\infty} |a_n| \) converges. This means that the series formed by taking the absolute values of the terms is convergent.

Step 2: Use Triangle Inequality

The triangle inequality states that for any real numbers \( a_1, a_2, \ldots, \) we have \( \left| \sum_{n=1}^{N} a_n \right| \leq \sum_{n=1}^{N} |a_n| \) for all natural numbers \( N \).

Step 3: Apply Limit to the Triangle Inequality

Since \( \sum_{n=1}^{\infty} |a_n| \) converges, for any \( \varepsilon > 0 \), there exists an \( N \) such that for all \( m > n \geq N \), the partial sum \( \sum_{k=n+1}^{m} |a_k| < \varepsilon \). Letting \( m \to \infty \), the series converges and we apply limits to the inequality: \( \left| \sum_{n=1}^{\infty} a_n \right| = \lim_{N \to \infty} \left| \sum_{n=1}^{N} a_n \right| \leq \lim_{N \to \infty} \sum_{n=1}^{N} |a_n| = \sum_{n=1}^{\infty} |a_n| \) which shows the required inequality.

Key Concepts

Triangle Inequality

The triangle inequality is a powerful tool in mathematics, particularly in the context of sequences and series. It states that for any finite sequence of real numbers, the absolute value of the sum is less than or equal to the sum of the absolute values. Formally, for a finite sequence \( a_1, a_2, \ldots, a_N \), the triangle inequality can be expressed as:

• \( \left| \sum_{n=1}^{N} a_n \right| \leq \sum_{n=1}^{N} |a_n| \)

This principle is significant because it allows us to estimate the magnitude of a series. In the context of absolute convergence, it ensures that even when the values of \( a_n \) are oscillating or alternating in sign, the absolute value of the entire sum cannot exceed the sum of the absolute individual terms.
This step is particularly useful when transitioning from finite sums to infinite series, a crucial aspect when establishing absolute convergence.

Convergence of Series

A series \( \sum_{n=1}^{\infty} a_n \) is said to converge when the sequence of its partial sums approaches a finite limit as more terms are added. There are two primary types of convergence to consider:

• Absolute Convergence: This occurs when the series of absolute values \( \sum_{n=1}^{\infty} |a_n| \) converges. Absolute convergence is a stronger condition than regular convergence because if a series converges absolutely, it converges in the standard sense as well.
• Conditional Convergence: This happens when a series \( \sum_{n=1}^{\infty} a_n \) converges, but the series of absolute values \( \sum_{n=1}^{\infty} |a_n| \) does not converge.

Establishing whether a series is absolutely convergent is crucial as it guarantees the usual properties of convergence, such as stability under reordering of terms.
For beginners, it's helpful to first check for absolute convergence, as it simplifies understanding and handling of infinite series.

Limit of Partial Sums

The concept of partial sums is pivotal in understanding series. For any series \( \sum_{n=1}^{\infty} a_n \), its corresponding partial sum is \( S_N = \sum_{n=1}^{N} a_n \). As \( N \to \infty \), if \( S_N \) converges to a specific number \( S \), we say the series converges to \( S \).

• If the absolute values of these partial sums \( \sum_{n=1}^{\infty} |a_n| \) converge, it indicates absolute convergence.
• This means that after adding many terms, the series settles down and approaches a predictable, finite limit.

The limit of partial sums is directly used in proofs and arguments where limits are applied to inequalities, such as the triangle inequality.
Understanding this concept aids in visualizing how adding more terms affects the sum and whether the entire series is bounded or diverging.