Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{k=1}^{\infty} \frac{-2}{\sqrt{k+2}}\)

Step-by-Step Solution

Answer

The series diverges.

Step 1: Analyze the Series

We need to evaluate the convergence of the series \( \sum_{k=1}^{\infty} \frac{-2}{\sqrt{k+2}} \). Since each term is negative, we consider the series \( \sum_{k=1}^{\infty} \frac{2}{\sqrt{k+2}} \) and determine if it converges.

Step 2: Use Direct Comparison Test

The terms \( \frac{2}{\sqrt{k+2}} \) are positive and we have a known series \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \), which diverges (p-series with \( p = 1/2 < 1 \)). We compare our series to this divergent series.

Step 3: Set Up Direct Comparison

Since \( \frac{2}{\sqrt{k+2}} > \frac{1}{\sqrt{k}} \) for all \( k \geq 1 \), by the direct comparison test, because the terms are larger than a divergent series, the series \( \sum_{k=1}^{\infty} \frac{2}{\sqrt{k+2}} \) also diverges.

Step 4: Conclusion on Convergence

Since \( \sum_{k=1}^{\infty} \frac{2}{\sqrt{k+2}} \) diverges, the original series \( \sum_{k=1}^{\infty} \frac{-2}{\sqrt{k+2}} \) also diverges because multiplying a divergent positive series by a negative constant results in the same divergence behavior.

Key Concepts

Direct Comparison Test

The direct comparison test is a valuable tool for determining whether a series converges or diverges. In essence, this test involves comparing a given series to another series whose convergence properties are already known. If the terms of the series you're examining are larger than a divergent series or smaller than a convergent series, you can infer the behavior of your original series.

Here's how it works:

• Start by identifying a series with known convergence or divergence.
• Compare the terms of your series with those of the known series.
• If each term of your series is less than or equal to a converging series term, your series converges.
• If each term of your series is greater than or equal to a diverging series term, your series diverges.

In our problem, we compared the terms of the series \( \sum_{k=1}^{\infty} \frac{2}{\sqrt{k+2}} \) with \( \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} \), which we know diverges. Since our series' terms are greater, it too diverges.

p-Series

A p-series is a special type of series defined as \( \sum_{k=1}^{\infty} \frac{1}{k^p} \) where \( p \) is a positive constant. The convergence or divergence of a p-series depends primarily on the value of \( p \).

Important points about p-series:

• The series converges if \( p > 1 \).
• The series diverges if \( p \leq 1 \).

In the context of our exercise, we dealt with a series similar to a p-series with \( p = 1/2 \). This case diverges because \( p \leq 1 \). By identifying this behavior, we can use it as a benchmark for comparing other series, as done in the exercise using the direct comparison test.

Divergence of Series

Determining whether a series diverges is a crucial aspect of series analysis, especially when dealing with infinite sums. Divergence means that the sum of the terms of the series increases without bound as more terms are added.

Key points to identify divergence:

• If the terms of a series do not approach zero, the series must diverge.
• If the series can be compared to another known divergent series where its terms are larger, it will also diverge (as in the direct comparison test).

In our exercise, once we established that the series \( \sum_{k=1}^{\infty} \frac{2}{\sqrt{k+2}} \) is divergent due to its comparison to a known p-series, the divergence of the original series \( \sum_{k=1}^{\infty} \frac{-2}{\sqrt{k+2}} \) is clear. The fact that it is multiplied by \(-2\) simply means the series' terms are negative, but this factor doesn't affect the divergence conclusion.