Problem 37 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Analyze Absolute Convergence

To determine absolute convergence, examine the series \( \sum_{n=1}^{\infty} \left| \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \right| = \sum_{n=1}^{\infty} \frac{(n+1)^{n}}{(2 n)^{n}} \). Simplify each term: \( \frac{(n+1)^{n}}{(2n)^{n}} = \left( \frac{n+1}{2n} \right)^{n} = \left( \frac{1}{2} + \frac{1}{2n} \right)^{n} \). As \( n \to \infty \), the expression approaches \( \left( \frac{1}{2} \right)^{n} \).

Step 2: Assess Series for Absolute Convergence

For large \(n\), the series \( \sum_{n=1}^{\infty} \left( \frac{1}{2} + \frac{1}{2n} \right)^{n} \approx \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^{n} \). This geometric series with a common ratio of \( r = \frac{1}{2} \) converges because \( |r| < 1 \). Therefore, the series \( \sum_{n=1}^{\infty} \left| \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \right| \) converges.

Step 3: Determine Conditional Convergence

Since \( \sum_{n=1}^{\infty} \left| \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \right| \) converges absolutely as determined in Step 2, it follows that the original series \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \) also converges. Absolute convergence implies convergence of the original alternating series.

Key Concepts

Absolute Convergence

When a series is said to converge absolutely, this means that if we take the absolute value of each term in the series and the resulting series still converges, then the original series also converges. This is a very strong form of convergence because it implies that even without considering the sign of each term, the series still adds up to a finite number.

Let's look at the series \[ \sum_{n=1}^{\infty} \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \]The absolute value series is\[ \sum_{n=1}^{\infty} \left| \frac{(-1)^{n}(n+1)^{n}}{(2 n)^{n}} \right| = \sum_{n=1}^{\infty} \frac{(n+1)^{n}}{(2 n)^{n}} \]By simplifying\[ \frac{(n+1)^{n}}{(2n)^{n}} = \left( \frac{1}{2} + \frac{1}{2n} \right)^{n} \]As \(n\) gets larger, the expression \(\left( \frac{1}{2} + \frac{1}{2n} \right)^{n}\) approaches \(\left( \frac{1}{2} \right)^{n}\). This resembles a geometric series with a common ratio of \(\frac{1}{2}\), which is less than 1, confirming that this series converges absolutely.

Alternating Series

Alternating series are series in which the terms alternate in sign, like positive, negative, positive, and so forth. Mathematically, they take the form \[\sum_{n=1}^{\infty} (-1)^{n} a_n\]where \(a_n\) is a positive sequence of numbers. Alternating series can converge even if they do not converge absolutely, thanks to a special test known as the Alternating Series Test.

This test states that an alternating series \(\sum (-1)^{n} a_n \) converges if:

• The sequence \(a_n\) is decreasing: \(a_{n+1} \leq a_n\) for all \(n\).
• The limit of \(a_n\) as \(n\) approaches infinity is zero: \(\lim_{n\to\infty} a_n = 0\).

In our case, since we already established that the series converges absolutely, it automatically converges according to the stronger absolute convergence criteria.

Geometric Series

A geometric series is a series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The general form of a geometric series is:\[ \sum_{n=0}^{\infty} ar^n\]where \(a\) is the first term and \(r\) is the common ratio.

A geometric series converges if the absolute value of the common ratio \(|r|\) is less than 1, and its sum is given by \[ \frac{a}{1-r}\]In the provided example, we examined the absolutes of terms and converted it into a geometric form: \[ \sum_{n=1}^{\infty} \left( \frac{1}{2} \right)^{n} \]Here, \(r = \frac{1}{2}\) satisfies \(|r| < 1\), showing convergence. This helps in establishing absolute convergence, which further corroborates that our series also behaves similarly to a geometric series over large values of \(n\), assuring it converges absolutely.