Question

State whether each of the following series converges absolutely, conditionally, or not at all\(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n+3}}{n}\)

Short answer

The series converges conditionally.

Step-by-step solution

Test for Absolutely Convergence

To check if the series converges absolutely, take the absolute value of the series: \[\sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{\sqrt{n+3}}{n}\right| = \sum_{n=1}^{\infty} \frac{\sqrt{n+3}}{n}\]Apply the Limit Comparison Test. Compare the given series with the divergent p-series \(\sum \frac{1}{\sqrt{n}}\).

Apply the Limit Comparison Test

We compare \(\frac{\sqrt{n+3}}{n}\) to \(\frac{\sqrt{n}}{n}\).Calculate the limit: \[\lim_{n \to \infty} \frac{\frac{\sqrt{n+3}}{n}}{\frac{\sqrt{n}}{n}} = \lim_{n \to \infty} \frac{\sqrt{n+3}}{\sqrt{n}}\]Simplify the expression:\[= \lim_{n \to \infty} \sqrt{\frac{n+3}{n}} = \lim_{n \to \infty} \sqrt{1 + \frac{3}{n}} = 1\]Since the limit is finite and non-zero, and \(\sum \frac{1}{\sqrt{n}}\) diverges, \(\sum \frac{\sqrt{n+3}}{n}\) also diverges.

Test for Conditional Convergence

Since the series does not converge absolutely, check for conditional convergence by applying the Alternating Series Test to \(\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\sqrt{n+3}}{n}\).To satisfy this test, the terms \(b_n = \frac{\sqrt{n+3}}{n}\) must be decreasing and the limit \(\lim_{n \to \infty} b_n = 0\) needs to hold.

Analyze for Decreasing Terms and the Limit Condition

For decreasing terms, consider:\[\frac{\sqrt{n+3}}{n} > \frac{\sqrt{n+4}}{n+1}\]After simplification, find that it holds for sufficiently large \(n\).Then, compute the limit condition:\[\lim_{n \to \infty} \frac{\sqrt{n+3}}{n} = \lim_{n \to \infty} \frac{\sqrt{n}}{n} = \lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0\]Since both conditions satisfy the Alternating Series Test, the series converges conditionally.

Key concepts

Alternating Series Test

The Alternating Series Test is an important tool when deciding on the convergence of series with terms that alternate in sign, like \((-1)^{n+1} a_n\). It primarily asks two key questions: Are the series terms decreasing in magnitude, and do they approach zero as \(n\) goes to infinity? In the given series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n+3}}{n}\), the test checks specifically the terms \(b_n = \frac{\sqrt{n+3}}{n}\).

• First, we ensure that \(b_n\) is decreasing. This is done by observing that \(\frac{\sqrt{n+3}}{n}\) becomes smaller as \(n\) increases because the numerator grows slower than the denominator.
• Second, we compute the limit of \(b_n\) as \(n\) approaches infinity. Calculating \(\lim_{n \to \infty} \frac{\sqrt{n+3}}{n}\) involves simplifying to \(\frac{1}{\sqrt{n}}\), giving a result of 0.

With these conditions satisfied, the series meets the criteria for the Alternating Series Test, indicating conditional convergence.

Absolute Convergence

Absolute convergence of a series refers to the situation where the series remains convergent even when all term signs are made positive. To test for absolute convergence, we typically examine the series formed by the positive magnitudes of its terms. For this series, \(\sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{\sqrt{n+3}}{n}\right|\), the problem simplifies to examining \(\sum_{n=1}^{\infty} \frac{\sqrt{n+3}}{n}\). To determine convergence, we applied the Limit Comparison Test.

• We compared this series to a known divergent p-series, \(\sum \frac{1}{\sqrt{n}}\).
• By calculating \( \lim_{n \to \infty} \frac{\frac{\sqrt{n+3}}{n}}{\frac{\sqrt{n}}{n}}\), which simplifies to \(1\), we conclude that the original series behaves similarly and thus diverges.

Since the series does not converge with positive terms, it does not converge absolutely.

Conditional Convergence

Conditional convergence describes a series that converges not by summing its absolute terms, but by leveraging alternating signs to achieve convergence. This type of convergence distinguishes itself by the failure of absolute convergence tests, implying a more delicate balance of terms.By analyzing the original series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n+3}}{n}\) using alternating signs, we confirmed conditional convergence through the Alternating Series Test:

• The solution stated \(b_n = \frac{\sqrt{n+3}}{n}\) is decreasing for sufficiently large \(n\) and its limit as \(n\) approaches infinity is 0.

The series, therefore, converges conditionally because its alternating nature is critical to convergence, while its absolute counterpart diverges. Understanding conditional convergence can often unravel surprising results for series that one might otherwise expect to not converge.