Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n(1+\sqrt{n})} $$

Step-by-Step Solution

Answer

The series is absolutely convergent.

Step 1: Identify the Alternating Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n(1+\sqrt{n})} \). This is an alternating series due to the factor \((-1)^{n+1}\), which causes the terms to alternate in sign.

Step 2: Test for Convergence of the Alternating Series

Apply the Alternating Series Test. For the series \( \sum (-1)^{n+1} a_n \), if \( a_n = \frac{1}{n(1+\sqrt{n})} \) is positive, decreasing, and \( \lim_{n \to \infty} a_n = 0 \), then the series converges.

Step 3: Check if the Series Terms are Positive

Clearly, \( \frac{1}{n(1+\sqrt{n})} > 0 \) for all \( n \geq 1 \). This satisfies the first condition of the Alternating Series Test.

Step 4: Check if Terms are Decreasing

Determine if \( a_{n+1} < a_n \). We calculate \( a_{n} = \frac{1}{n(1+\sqrt{n})} \) and \( a_{n+1} = \frac{1}{(n+1)(1+\sqrt{n+1})} \), and show that \( a_{n+1} < a_n \) using derivative or algebraic comparison for decreasing nature.

Step 5: Calculate the Limit of the Sequence

Compute \( \lim_{n \to \infty} \frac{1}{n(1+\sqrt{n})} \). As \( n \to \infty \), \( \sqrt{n} \) grows less quickly than \( n \), and the limit approaches 0, satisfying the third condition of the Alternating Series Test.

Step 6: Evaluate Absolute Convergence

Examine the absolute convergence using \( \sum_{n=1}^{\infty} \left| \frac{1}{n(1+\sqrt{n})} \right| \). Because this series behaves similarly to \( \frac{1}{n^{3/2}} \), it can be compared to a p-series with \( p = 3/2 > 1 \), which converges. Therefore, the series is absolutely convergent.

Key Concepts

Alternating Series Test

An alternating series is one in which the terms alternate in sign, typically due to a factor like \((-1)^{n+1}\). Think of it as flipping signs — positive to negative and back again. The series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n(1+\sqrt{n})}\) is a perfect example of an alternating series. To determine if it converges, we employ the Alternating Series Test.

This test requires three things from the series:

• The terms \(a_n = \frac{1}{n(1+\sqrt{n})}\) must be positive.
• The terms \(a_n\) should be decreasing.
• The limit of \(a_n\) as \(n\) approaches infinity should be zero.

For the series in question, with \(a_n = \frac{1}{n(1+\sqrt{n})}\): - As \(n\) gets larger, the terms decrease in size. - The limit approaches zero as \(n\) goes to infinity.

So, we pass all three requirements, confirming the series converges!

Absolute Convergence

Absolute convergence is when a series of absolute values converges.

Even if the series oscillates, like our alternating series, it is examined using the absolute terms. To find this out, we look at \(|a_n|\) — that’s the magnitude of terms without the signs flipping.

For our series: - The absolute series is \(\sum_{n=1}^{\infty} \left| \frac{1}{n(1+\sqrt{n})} \right|\). This behaves similarly to \(\frac{1}{n^{3/2}}\), a p-series.

Since \(p = 3/2 > 1\), and a known p-series with \(p > 1\) converges, the absolute series here converges too. When this absolute series converges, the original alternating series isn’t just convergent, it’s absolutely convergent!

Conditional Convergence

A series is conditionally convergent if the series converges, but not when you take the absolute values of its terms.

This happens when only some parts or conditions ensure convergence, maybe because of the alternating nature of the series rather than straightforward number sums converging.

But, remember: if a series like our given one can pass the absolute convergence test, then it's not just conditionally convergent anymore; it's absolutely convergent! We explored how the absolute form of our series converged, hence, it's not merely conditionally convergent. Conditional convergence steps back when absolute convergence steps in.

P-Series Comparison

The p-series is one where each term is of the form \(\frac{1}{n^p}\). It's a classic benchmark to see if other series converge. That's what we call a p-series comparison!

For convergence:

• If \(p > 1\), it converges.
• If \(p \leq 1\), it diverges.

For our given series with absolute terms \(\left| \frac{1}{n(1+\sqrt{n})} \right|\), we checked it behaves like \(\frac{1}{n^{3/2}}\), a p-series where \(p = 3/2\). Since \(p\) is more than 1, it converges, ensuring the absolute convergence of the series we analyzed.

Using p-series comparison gives a straightforward path to check for convergence, making it easier when facing complex series behavior!