Problem 41 - Calculus: Early Transcendentals

Question

Let \( \\{ {b_n} \\} \) be a sequence of positive numbers that converges to \( \frac {1}{2}. \) Determine whether the given series is absolutely convergent. \( \displaystyle \sum_{n = 1}^{\infty} \frac {{{b_{n }^{n} \cos n \pi }}}{n} \)

Step-by-Step Solution

Answer

The series is absolutely convergent.

Step 1: Analyze the General Term

First, we examine the general term of the series, which is \( a_n = \frac{b_n^n \cos(n\pi)}{n} \). Since \( \cos(n\pi) = (-1)^n \), the term simplifies to \( a_n = \frac{b_n^n (-1)^n}{n} \). To determine absolute convergence, we need to analyze \( |a_n| = \frac{|b_n^n|}{n} \), since \(|(-1)^n|=1\).

Step 2: Evaluate the Limit of the General Term

Evaluate \(|b_n^n|\) as \(n \to \infty\). We know that \(b_n \to \frac{1}{2}\) as \(n \to \infty\). So, we can approximate \(b_n \approx \frac{1}{2}\) for large \(n\). Thus, \(b_n^n \approx \left(\frac{1}{2}\right)^n\).

Step 3: Determine the Convergence of \(|a_n|\)

Substitute the approximation into \(|a_n| = \frac{b_n^n}{n}\), yielding \( |a_n| \approx \frac{\left(\frac{1}{2}\right)^n}{n} \). Use the fact that \(\sum \frac{(\frac{1}{2})^n}{n}\) is a convergent series because it is a series of a geometric sequence \(\left(\frac{1}{2}\right)^n\) multiplied by terms \(\frac{1}{n}\) which decrease faster than harmonic series. Thus, \(|a_n|\to 0\) as \(n \to \infty\).

Step 4: Conclude Absolute Convergence

Since \(\sum \frac{(\frac{1}{2})^n}{n}\) is convergent, by comparison, \(\sum |a_n|\) is also convergent. Therefore, the series \(\sum a_n\) is absolutely convergent as \(|a_n| \approx \frac{(\frac{1}{2})^n}{n}\) is a convergent series due to the rapid decay of the terms.

Key Concepts

Sequence Convergence

In mathematics, sequence convergence describes the behavior of a sequence as it approaches a specific value as the number of terms goes to infinity. For a sequence \( \{ b_n \} \) converging to \( \frac{1}{2} \), it means that as \( n \) becomes very large, the terms \( b_n \) get closer and closer to \( \frac{1}{2} \).
This concept is essential as it allows mathematicians to make predictions about the behavior of sequences and functions over time.
Convergence indicates stability, implying that the sequence does not diverge to infinity or oscillate wildly.
This becomes crucial in series analysis where the limit of the sequence's terms directly affects the convergence of the series.

Geometric Series

A geometric series is a series with a constant ratio between successive terms, defined generally as \( ar^0 + ar^1 + ar^2 + \, ... \). A famous geometric series example is \( \sum_{n=0}^{\infty} ar^n \). The series converges if the absolute value of the ratio \( |r| < 1 \) and its sum is \( \frac{a}{1-r} \).
In the given problem, the term \( \left(\frac{1}{2}\right)^n \) resembles a geometric series since each term is a constant multiple of the previous term. Due to the term \( \frac{1}{n} \) accompanying \( \left(\frac{1}{2}\right)^n \) in this exercise, the series combines both geometric and harmonic characteristics.
This helps in showing convergence by comparing with a basic geometric series.

Harmonic Series

The harmonic series is the sum \( \sum_{n=1}^{\infty} \frac{1}{n} \), notorious for its tendency to diverge, meaning it increases indefinitely without bound.
Despite its divergence, the harmonic series decreases slowly, which is crucial when combined with other factors in series tests.
In the analyzed series, the general term is \( \frac{\left(\frac{1}{2}\right)^n}{n} \), where \( \frac{1}{n} \) has harmonic characteristics. However, its fast decay, due to \( \left(\frac{1}{2}\right)^n \), makes it converge rather than diverge.
This unique blend accelerates the convergence of the series, as the geometric factors significantly reduce the impact of the harmonic terms.

General Term Analysis

Analyzing the general term of a series involves assessing its components to determine overall series behavior.
In our series, the general term is \( |a_n| = \frac{|b_n|^n}{n} \), simplifying to \( \frac{\left(\frac{1}{2}\right)^n}{n} \) due to the convergence of \( b_n \) to \( \frac{1}{2} \).
Such complex expressions involve breaking them into simpler forms to compare against well-known convergent or divergent series.
This step determines if a series will converge absolutely, conditionally, or not at all.
Through thorough examination, like comparing with a convergent series \( \sum \frac{1}{2^n} \), the larger series is determined as absolutely convergent, settling the analysis conclusively.