Problem 27 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n} n^{2}(2 / 3)^{n} $$

Step-by-Step Solution

Answer

The series converges absolutely because the series with absolute terms \( \sum_{n=1}^{\infty} n^{2}(2 / 3)^{n} \) converges.

Step 1: Identify the Absolute Series

The given series is \[ \sum_{n=1}^{\infty}(-1)^{n} n^{2}(2 / 3)^{n}. \]To determine absolute convergence, we first consider the absolute value of the series terms:\[ \sum_{n=1}^{\infty} n^{2}(2 / 3)^{n}. \]

Step 2: Test Absolute Convergence using the Ratio Test

We will use the Ratio Test to check for absolute convergence.The Ratio Test states that for a series \( \sum a_n \), if \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), then the series converges absolutely.Let \( a_n = n^2 \left( \frac{2}{3} \right)^n \). Calculate:\[\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 \left( \frac{2}{3} \right)^{n+1}}{n^2 \left( \frac{2}{3} \right)^n} = \frac{(n+1)^2}{n^2} \cdot \frac{2}{3}.\]Simplifying gives:\[\frac{(n+1)^2}{n^2} \cdot \frac{2}{3} \approx \left( 1 + \frac{2}{n} + \frac{1}{n^2} \right) \cdot \frac{2}{3}.\]Taking the limit as \( n \to \infty \):\[L = \lim_{n \to \infty} \left( \frac{2}{3} \cdot \left( 1 + \frac{2}{n} + \frac{1}{n^2} \right) \right) = \frac{2}{3}.\]Since \( L < 1 \), the series \( \sum_{n=1}^{\infty} n^{2}(2 / 3)^{n} \) converges absolutely.

Step 3: Consider the Original Series with Alternating Terms

Since the absolute convergence of \( \sum_{n=1}^{\infty} n^{2}(2 / 3)^{n} \) is established, the original alternating series \( \sum_{n=1}^{\infty}(-1)^{n} n^{2}(2 / 3)^{n} \) also converges.

Key Concepts

Absolute Convergence

When we talk about absolute convergence in a series, we mean that if you take the absolute values of each term and then sum them up, it still converges. Think of it like measuring the series without the negative signs. The original problem involves handling the series \[ \sum_{n=1}^{\infty}(-1)^{n} n^{2}(2 / 3)^{n} \]. To check for absolute convergence, we eliminate the alternating negative and positive signs by considering only \[ \sum_{n=1}^{\infty} n^{2}(2 / 3)^{n}. \]**Why Check Absolute Convergence First?**- If this "absolute" series converges, the original alternating series also converges.- Absolute convergence implies that the series is very, very "stable," losing out on divergence even if terms change signs.In this case, because the "absolute" series converges, so does our original series. This indicates that even if each term's sign flips, the series is plentifully tame to settle down into a finite value.

Alternating Series

Alternating series are those that switch signs each term, like our original series \[ \sum_{n=1}^{\infty}(-1)^{n} n^{2}(2 / 3)^{n}. \]In this series, the powers of \(-1\) cause the sign changes because \((-1)^n\) is positive for even \(n\) and negative for odd \(n\).**Why Use Alternating Series?**- They are helpful for approximations in mathematics.- Only special kinds of alternating series may converge (those where terms themselves go to zero and are in a bouncing sort of approach to zero).**Characteristics of Converging Alternating Series:**- Terms reduce in size- The positive and negative balancing helps settle towards a sumOnce we confirmed the absolute convergence, as in this series, we can say that \[ \sum_{n=1}^{\infty}(-1)^{n} n^{2}(2 / 3)^{n} \] converges, making it well-behaved and revealing that those sign changes alone don't push it towards divergence.

Ratio Test

The Ratio Test is a powerful tool to determine if a series absolutely converges. It involves creating a limit that compares the size of consecutive series terms. Here, \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] where \(a_n = n^2 (2/3)^n\) for the absolute series.**When Applying the Ratio Test:**- Calculate \(\frac{a_{n+1}}{a_n}\) by substituting \(n + 1\) into your term and dividing by the \(n\)th term.- Simplify this expression so that you can see what happens as \(n\) becomes infinitely large.- Take the limit; if \(L < 1\), convergence is assured.**Outcome of Ratio Test:**- For \(L = \frac{2}{3}\), which is less than 1, our series converges absolutely.- The test's result being less than 1 confirms both the workability for infinite terms and strengthens our affirmation that the alternating series is entirely under control.Using the Ratio Test in cases like this is quicker and highly insightful, providing a clear path to understanding a series' behavior.