Problem 42 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The series converges but does not converge absolutely.

Step 1: Estimate the General Term

Start by analyzing the general term of the series: \[ a_n = (-1)^n \left(\sqrt{n^2+n} - n\right). \]The term can be rewritten as \[ a_n = (-1)^n \left(\frac{n^2+n}{\sqrt{n^2+n}+n}\right). \]Simplifying further using rationalization or approximations, you get the leading order term:\[ a_n \approx (-1)^n \left(\frac{1}{2}\sqrt{\frac{1}{n}}\right). \]

Step 2: Apply the Alternating Series Test

The alternating series test states that an alternating series \( \sum (-1)^n b_n \) converges if the sequence \( b_n \) is decreasing and \( \lim_{n \to \infty} b_n = 0 \).Here, \( b_n = \frac{1}{2}\sqrt{\frac{1}{n}} \), which is decreasing and approaches 0 as \( n \to \infty \). Thus, the series converges.

Step 3: Check for Absolute Convergence

To check for absolute convergence, consider the series of absolute values: \[ \sum \left| (-1)^n \left(\sqrt{n^2+n} - n\right) \right| = \sum \frac{1}{2}\sqrt{\frac{1}{n}}. \]This is a form similar to the \( p \)-series with \( p = \frac{1}{2} \), which diverges because \( p \leq 1 \). Thus, the series does not converge absolutely.

Key Concepts

Alternating Series Test

The Alternating Series Test is a crucial tool when examining series with alternating signs. When faced with a series of the form \( \sum (-1)^n b_n \), to determine convergence, you need to investigate two conditions.

• First, ensure that the sequence \( b_n \) is decreasing as \( n \) increases. This means that each term in the sequence is smaller than the previous term.
• Second, check that the limit of \( b_n \) as \( n \) approaches infinity is zero: \( \lim_{n \to \infty} b_n = 0 \).

If both these criteria are satisfied, the alternating series converges. In our original exercise, \( b_n = \frac{1}{2}\sqrt{\frac{1}{n}} \), which is indeed a decreasing sequence. Also, since \( \lim_{n \to \infty} \frac{1}{2}\sqrt{\frac{1}{n}} = 0 \), the series is confirmed to converge by the Alternating Series Test. This tool provides a straightforward approach to understanding particular alternating series and their behaviors.

Absolute Convergence

Absolute convergence examines if the series of absolute values converges. This is a stronger condition than regular convergence. When analyzing absolute convergence for a series \( \sum a_n \), evaluate the series \( \sum |a_n| \). If \( \sum |a_n| \) converges, then \( \sum a_n \) converges absolutely. However, if the series of absolute values diverges, the original series can still converge, but not absolutely. In the step by step solution, the series \( \sum \left| (-1)^n \left(\sqrt{n^2+n} - n\right) \right| = \sum \frac{1}{2}\sqrt{\frac{1}{n}} \) is checked. Since this is similar to a \( p \)-series with \( p = \frac{1}{2} \), and known to diverge because \( p \leq 1 \), it means the original series does not converge absolutely. Recognizing when a series doesn't converge absolutely is vital for understanding the limits of its convergence.

p-Series

The concept of a \( p \)-series is fundamental in studying series convergence. A typical \( p \)-series is expressed as \( \sum \frac{1}{n^p} \), where \( p \) is a positive constant.

• The convergence or divergence of a \( p \)-series is directly determined by the value of \( p \).
• If \( p > 1 \), the \( p \)-series converges.
• If \( p \leq 1 \), the series diverges.

It's essential to identify when a series resembles a \( p \)-series as it can simplify the convergence analysis. In our example, after simplification, the terms looked similar to a \( p \)-series with \( p = \frac{1}{2} \). Because \( p = \frac{1}{2} \leq 1 \), the conclusion was drawn that this related series diverges, impacting the discussion on absolute convergence of the original series. Understanding the nature of \( p \)-series helps in quickly assessing many series you might encounter.