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Chapter 10: Problem 37

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}}$$

Short Answer

Expert verified
The series diverges, neither converging absolutely nor conditionally.

Step by step solution

01

Understand Absolute Convergence

A series \( \sum a_n \) converges absolutely if \( \sum |a_n| \) converges. First, let's examine the absolute value of the series term: \(|a_n| = \left| \frac{(-1)^{n}(n+1)^{n}}{(2n)^{n}} \right| = \frac{(n+1)^{n}}{(2n)^{n}} \).
02

Simplify the Series Term

We simplify \( \frac{(n+1)^{n}}{(2n)^{n}} \) as follows:\[ \left(\frac{n+1}{2n} \right)^{n} = \left(\frac{n+1}{n} \cdot \frac{1}{2} \right)^{n} = \left(1 + \frac{1}{n} \right)^{n} \cdot \left(\frac{1}{2} \right)^{n} \].The expression \( \left(1 + \frac{1}{n} \right)^{n} \) tends to \(e\) as \( n \) approaches infinity, because it approximates the exponential function. So:\[ \left(\frac{(n+1)^{n}}{(2n)^{n}} \right) \approx \left(\frac{e}{2} \right)^{n} \].
03

Check Absolute Convergence

Consider the series \( \sum \left(\frac{e}{2}\right)^n \). Note that \( \frac{e}{2} > 1 \) because \( e \approx 2.718 \). Therefore, \( \left(\frac{e}{2}\right)^n \) is exponentiating a value greater than 1, which makes the absolute series diverge.
04

Check Conditional Convergence

The series given is \( \sum (-1)^n \left(\frac{(n+1)^{n}}{(2n)^{n}} \right) \), an alternating series. For an alternating series \( \sum (-1)^n b_n \) to converge, it needs \( b_n \to 0 \) and \( b_n \) to be eventually decreasing. Given the earlier simplification to \( \left(\frac{e}{2}\right)^n \), and since \( \left(\frac{e}{2}\right)^n \) diverges and doesn't approach zero, the original series diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Convergence
When we talk about absolute convergence of a series, we are interested in whether the series remains convergent when we substitute each term with its absolute value. **In simpler terms:**
  • If \( \sum |a_n| \) converges, then \( \sum a_n \) also converges absolutely.
  • Absolute convergence is a very strong form of convergence.
For example, in the exercise, the term \( a_n = \frac{(-1)^{n}(n+1)^{n}}{(2n)^{n}} \) was checked for absolute convergence by looking at its absolute form \( |a_n| = \frac{(n+1)^{n}}{(2n)^{n}} \). Simplifying, we found that the sequence behaves like \( \left(\frac{e}{2}\right)^n \) for large \(n\).
Because \( \frac{e}{2} > 1 \), the absolute series \( \sum \left(\frac{e}{2}\right)^n \) diverges. That means, the given series does not converge absolutely.
Conditional Convergence
Conditional convergence is slightly tricky, but worth understanding. This type of convergence occurs when a series converges, but its absolute version does not. For an **intuitive grasp**:
  • The series itself converges.
  • However, the absolute value series diverges.
In the solution step provided, the series \( \sum (-1)^n \left(\frac{(n+1)^{n}}{(2n)^{n}} \right) \) was looked at for conditional convergence. Since its absolute form diverged and as the terms do not approach zero (\( b_n = \left(\frac{e}{2}\right)^n \)), it can't be conditionally convergent either. The series itself thus does not converge conditionally.
Alternating Series
An alternating series is a fascinating concept where terms alternate in sign. **Key aspects include:**
  • Format: It usually looks like \( \sum (-1)^n b_n \).
  • Behavior: The terms switch between positive and negative.
For convergence, such a series requires its absolute term to approach zero as \(n\) increases, and the terms must be eventually decreasing.
The exercise example, \( \sum (-1)^n \left(\frac{(n+1)^{n}}{(2n)^{n}} \right) \), fits the alternating series format but does not satisfy the convergence criteria. Here, because the simplified term \( \left(\frac{e}{2}\right)^n \) does not tend to zero and isn't decreasing, it confirms the series' divergence.
Divergence in Sequences
Divergence refers to a series or sequence that does not settle or approach a definite limit. In the world of sequences and series:
  • If terms don't get smaller and close to zero, the series will diverge.
  • Divergence indicates lack of convergence.
In our solved problem, upon stripping the signs and exploring \( \sum \left(\frac{e}{2}\right)^n \), we determined that it grows without bound because \( \frac{e}{2} > 1 \).
This is the hallmark of divergence鈥� the terms are increasing rather than fading away to zero. Thus, understanding divergence helps identify where a series or sequence will falter in becoming convergent.

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