Problem 46 Determine whether the series is ... [FREE SOLUTION]

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Calculus

Deborah Hughes-Hallett, Andrew M. Gleason, William G. McCallum

$Math Studyset 91影视 Explanations$ Math

7 Edition

Chapter 9: Problem 46

Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$$

Short Answer

Expert verified

The series is conditionally convergent.

Step by step solution

Identify the Series Type

The series given is $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$. It is an alternating series because of the term $(-1)^{n-1}$, which means the signs of the terms alternate between positive and negative.

Check for Absolute Convergence

To determine absolute convergence, we consider the series $\sum_{n=1}^{\infty} \left| \frac{(-1)^{n-1}}{\sqrt{n}} \right| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$. This is a p-series with $p = \frac{1}{2}$. A p-series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$. Since $\frac{1}{2} < 1$, the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges, thus the original series does not converge absolutely.

Check for Conditional Convergence

The series meets the conditions of the Alternating Series Test: The terms $\frac{1}{\sqrt{n}}$ are positive, decreasing, and their limit as $n \to \infty$ is 0. Thus, the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$ converges conditionally.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Convergence

Absolute convergence is a term used when examining a series to determine if it converges when all terms are made positive. For the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$, absolute convergence is checked by looking at $\sum_{n=1}^{\infty} \left| \frac{(-1)^{n-1}}{\sqrt{n}} \right| $, which simplifies to $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$. This series transformation allows us to see if everything still sums up to a finite number when keeping only positive terms.

If a series is absolutely convergent, then it converges no matter how the terms' signs are changed.
Absolute convergence implies both convergence of the series itself and the stabilized sum of its terms.

For our given series, this transformed series is identified as a p-series with $p = \frac{1}{2}$, which does not meet the condition for absolute convergence, causing it instead to diverge.

Conditional Convergence

Conditional convergence occurs when a series converges, but does not converge absolutely. This means the original series adds up to a finite sum only because of the alternating positive and negative terms, which results in partial cancellation.In the case of our series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$, although we found it does not converge absolutely, it does converge conditionally. This is proven using the Alternating Series Test:

The terms $\frac{1}{\sqrt{n}}$ are positive and decrease as $n$ increases.
The limit of these terms as $n$ approaches infinity is zero.

Since these conditions are met, the series is indeed conditionally convergent, suggesting that sign alternation plays a crucial role in making the series convergent.

Divergence

Divergence is the opposite of convergence in series. When a series diverges, its terms do not settle towards a single finite sum; instead, they grow without bound or oscillate indefinitely. In the context of absolute convergence, our series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ provides a powerful demonstration of divergence:

This series is identified as a p-series with the exponent $p$ less than 1 (specifically, $p = \frac{1}{2}$).
P-series are known to diverge when $p \leq 1$.

Thus, while the alternating series converges conditionally, the series of absolute values does not, highlighting its divergence.

p-Series

A p-series is a specific type of series in mathematics of the form $\sum_{n=1}^{\infty} \frac{1}{n^p}$. The parameter $p$ plays a crucial role in determining whether the series converges or diverges:

If $p > 1$, the p-series converges, meaning the infinite sum approaches a finite value.
If $p \leq 1$, the series diverges, indicating the sum stretches to infinity.

The critical distinction arises because of how terms shrink as $n$ becomes large. In the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$, since $p = \frac{1}{2}$, the series diverges as $\frac{1}{2} < 1$. Understanding p-series provides clarity in solving such problems, and it is a primary tool in assessing series' behavior.

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91影视

Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n}}$$

Short Answer

Step by step solution

Identify the Series Type

Check for Absolute Convergence

Check for Conditional Convergence

Key Concepts

Absolute Convergence

Conditional Convergence

Divergence

p-Series

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