Problem 6 - Basic Analysis: Introduction to Real Analysis

Question

Prove the following stronger version of the ratio test: Let \(\sum x_{n}\) be a series. a) If there is an \(N\) and a \(\rho<1\) such that for all \(n \geq N\) we have \(\frac{x_{n+1}}{x_{n}}<\rho,\) then the series converges absolutely. b) If there is an \(N\) such that for all \(n \geq N\) we have \(\frac{\left|x_{n+1}\right|}{x_{n}} \geq 1,\) then the series diverges.

Step-by-Step Solution

Answer

If \(\frac{x_{n+1}}{x_{n}}<\rho<1\) for \(n\geq N\), the series converges absolutely; if \(\frac{\left|x_{n+1}\right|}{x_{n}} \geq 1\), it diverges.

Step 1: Understanding the Condition for Convergence

The condition given is that for all \(n \geq N\), \(\frac{x_{n+1}}{x_{n}}<\rho\), where \(\rho<1\). This means that each term in the sequence \(x_n\) is smaller than its predecessor by a constant ratio \(\rho\).

Step 2: Applying the Convergence Criteria

Since \(\rho<1\), the terms \(x_{n}\) are decreasing geometrically. The geometric series \(\sum x_{n}\) formed will converge absolutely because the terms become smaller and smaller as \(n\) increases.

Step 3: Proving Absolute Convergence

Consider the terms \(\left|x_{n}\right|\), they satisfy \(\left|x_{n+1}\right| < \rho \cdot \left|x_{n}\right|\). Thus, the series \(\sum \left|x_{n}\right|\) is bounded above by a convergent geometric series with ratio \(\rho<1\). Hence, the series converges absolutely.

Step 4: Understanding the Condition for Divergence

The condition given is that for all \(n \geq N\), \(\frac{\left|x_{n+1}\right|}{x_{n}} \geq 1\). This implies that the sequence terms \( |x_n| \) do not decrease.

Step 5: Applying the Divergence Criteria

If the terms \( |x_n| \) remain the same or increase, given enough terms, \( \sum |x_n| \) cannot converge. Thus, the series \(\sum x_{n}\) must diverge based on this criterion.

Key Concepts

Ratio Test

The Ratio Test is a popular method for determining the convergence or divergence of an infinite series. It involves examining the limit of the ratio of consecutive terms in the series. The test states that for a series \( \sum x_{n} \):

• If \( \lim_{{n \to \infty}} \left| \frac{x_{n+1}}{x_{n}} \right| = L \) and \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L = \infty \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

In the exercise, a stronger version of this test is applied. When there exists a \( \rho < 1 \) such that \( \frac{x_{n+1}}{x_{n}} < \rho \) for all \( n \geq N \), the series not only converges, but it converges absolutely. This additional requirement strengthens the predictive power of the Ratio Test.

Geometric Series

A geometric series is a series of the form \( \sum ar^{n} \), where \( a \) is the first term and \( r \) is the common ratio. The series converges if the absolute value of the common ratio \( |r| < 1 \), and its sum can be calculated using the formula:

• \( S = \frac{a}{1 - r} \)

If \( |r| \geq 1 \), the series diverges. In the context of the problem, when each term \( x_{n} \) satisfies \( \frac{x_{n+1}}{x_{n}} < \rho \) (where \( \rho < 1 \)), the series behaves similarly to a geometric series. Hence, it converges absolutely because the terms decrease geometrically by the factor of \( \rho \).

Absolute Convergence

Absolute convergence refers to a condition where the series \( \sum x_{n} \) converges even when each term is replaced by its absolute value \( \sum |x_{n}| \). If a series converges absolutely, it also converges in the regular sense. That is, the series \( \sum x_{n} \) and \( \sum |x_{n}| \) both have finite sums.

• When applying the first part of the stronger ratio test, we demonstrate absolute convergence by showing that \( \sum |x_{n}| \) is bounded by a convergent geometric series.
• This ensures the sequence terms become smaller over time, ensuring convergence without regard to sign changes.

Understanding absolute convergence helps in assessing the behavior of series, particularly when dealing with complex series where term signs could fluctuate.

Divergence

Divergence occurs when a series \( \sum x_{n} \) does not converge to a finite sum. In the exercise, if \( \frac{\left| x_{n+1} \right|}{x_{n}} \geq 1 \) for all \( n \geq N \), each term \( |x_{n}| \) does not decrease.

• This leads to a situation where the sums grow indefinitely and the series cannot approach a finite limit.
• The condition ensures that the positive sequence terms either remain constant or greater, meaning their sum cannot settle to a finite value.
• Hence, under this criterion, the series diverges.

Recognizing divergence is crucial in series analysis, as it dictates that efforts to find a finite sum are futile given the series' nature.