Chapter 11: Problem 41
Let \(\left\\{b_{n}\right\\}\) be a sequence of positive numbers that converges to \(\frac{1}{2} .\) Determine whether the given series is absolutely convergent. $$ \sum_{n=1}^{\infty} \frac{b_{n}^{n} \cos n \pi}{n} $$
Short Answer
Expert verified
The series is absolutely convergent.
Step by step solution
01
Understand Absolute Convergence
A series \(\sum a_n\) is absolutely convergent if \(\sum |a_n|\) converges. This means we need to consider the series \(\sum \left|\frac{b_{n}^{n} \cos n \pi}{n}\right|\).
02
Simplify the Absolute Value
Since \(\cos n\pi\) alternates between \(-1\) and \(1\), \(|\cos n\pi| = 1\) for all \(n\). This simplifies the series to \(\sum \frac{|b_n|^n}{n}\) because \(b_n\) is positive, \(|b_n| = b_n\).
03
Consider Limit of \(b_n^n\)
Given \(b_n \to \frac{1}{2}\) as \(n \to \infty\), we evaluate \(b_n^n = \left(\frac{1}{2} + \epsilon_n\right)^n\), where \(\epsilon_n \to 0\). Since \(\left(\frac{1}{2} + \epsilon_n\right)^n\) behaves like an exponential \(\left(\frac{1}{2}\right)^n\), this tends exponentially to zero.
04
Apply Comparison Test
Notice the expression \(\frac{|b_n|^n}{n}\). If \(b_n^n\) behaves like \(\left(\frac{1}{2}\right)^n\), then a comparison with \(\sum \frac{1}{2^n}{n}\) is logical. The series \(\frac{1}{2^n}{n}\) converges since \(\sum \frac{1}{2^n}\) converges (geometric series) and division by \(n\) ensures convergence.
05
Conclusion Using Comparison
Since \(\frac{b_n^n}{n} \leq \frac{C}{2^n}\) for some constant \(C > 0\) and \(\sum \frac{C}{2^n}\) converges, by the Comparison Test, \(\sum \frac{b_n^n}{n}\) also converges. Thus, \(\sum \frac{b_n^n \cos n \pi}{n}\) is absolutely convergent.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Convergence
In mathematics, absolute convergence is a concept where the absolute value of each term in a series is taken into account to determine convergence. A series \(\sum a_n\) is said to be absolutely convergent if the series of absolute values \(\sum |a_n|\) converges. This implies that if you strip away the signs of each term and form a new series with those values, the summation of that series must also reach a finite value.
- Absolute convergence implies regular convergence, but the reverse is not always true.
- This ensures that rearranging terms doesn't affect the sum, a quite useful property.
Comparison Test
The comparison test is a technique used for analyzing the convergence or divergence of infinite series or sequences. It involves comparing a given series with another series whose convergence behavior is known.
- If a series \(\sum a_n\) is given, and there exists another series \(\sum b_n\) such that \(0 \leq a_n \leq b_n\), the convergence of \(\sum b_n\) implies the convergence of \(\sum a_n\).
- The opposite holds for divergence as well; if \(\sum b_n\) diverges, then so does \(\sum a_n\) given \(a_n \geq b_n \geq 0\).
Sequences of Numbers
Sequences are ordered lists of numbers following a particular rule or pattern. When dealing with infinite sequences, understanding their limit behaviors is crucially important, especially for convergence investigations.
- A sequence \(\left\{b_n\right\}\) with a limit \(\frac{1}{2}\) means that as \(n\) grows larger, \(b_n\) gets closer and closer to \(\frac{1}{2}\).
- Limits are frequently used in describing the long-term behavior of sequences, which is critical in evaluating series.
Cosine Function
The cosine function is a periodic trigonometric function expressed as \(\cos x\). It's periodic with a period of \(2\pi\), meaning it repeats every \(2\pi\) radians or 360 degrees.
- For integer multiples \(n\), the value of \(\cos n\pi\) alternates between \(-1\) and \(1\).
- This periodicity directly affects the behavior of terms in trigonometric series, causing them to oscillate.