Question

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

Short answer

The series converges conditionally.

Step-by-step solution

Identify Series Type

The series given is \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{\sqrt{n+3}}\). This is an alternating series because of the term \((-1)^{n+1}\). To check for convergence, we will use the Alternating Series Test.

Apply Alternating Series Test

The Alternating Series Test states that an alternating series \(\sum (-1)^{n} a_n\) converges if \(a_n\) is positive, decreasing, and \(\lim_{n \to \infty} a_n = 0\). In this series, \(a_n = \frac{1}{\sqrt{n+3}}\), which is positive, and it decreases as \(n\) increases. The limit \(\lim_{n \to \infty} \frac{1}{\sqrt{n+3}} = 0\). Thus, the series converges.

Check for Absolute Convergence

To check for absolute convergence, consider the series \(\sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{1}{\sqrt{n+3}} \right| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+3}}\). Use the p-test for convergence with \(p = \frac{1}{2}\), where \(p < 1\), indicating the series diverges. Thus, the series does not converge absolutely.

Conclusion on Convergence Type

Since the original series converges by the Alternating Series Test but does not converge absolutely, it converges conditionally.

Key concepts

Alternating Series Test

An alternating series is one where the terms alternate in sign, such as \((-1)^{n+1}\). This test helps us determine whether such a series converges. For an alternating series \(\sum (-1)^n a_n\), to converge, it must satisfy three criteria:

• Each term \(a_n\) must be positive.
• The sequence \(a_n\) must be decreasing; that is, \(a_{n+1} < a_n\).
• The limit of \(a_n\) as \(n\) approaches infinity must be zero.

In the given problem, \(a_n = \frac{1}{\sqrt{n+3}}\), which is clearly positive for all \(n\). It also decreases with increasing \(n\) because larger denominators in fractions lead to smaller values. Finally, the limit goes to zero, \(\lim_{n \to \infty} \frac{1}{\sqrt{n+3}} = 0\). Thus, our series converges according to the Alternating Series Test.

Absolute Convergence

Absolute convergence is a stricter form of convergence, where a series of terms converges when all its terms are made positive. To check for absolute convergence in the given series, \(\sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{1}{\sqrt{n+3}} \right|\), we consider the sum \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+3}}\). Now, we apply the comparison test to determine if this series converges. Since the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+3}}\) resembles the harmonic series \(\sum \frac{1}{n^p}\) with \(p = \frac{1}{2}\), which diverges as \(p < 1\), the original series does not converge absolutely. Absolute convergence is vital since it guarantees both types of series (alternating and non-alternating) will sum to a finite value, regardless of the arrangement of terms.

Conditional Convergence

Conditional convergence occurs when a series converges according to the Alternating Series Test but not absolutely. Our original alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{\sqrt{n+3}}\) meets the criteria for convergence using the Alternating Series Test, as discussed before. However, when we examined whether it converges absolutely, we found that the series of the absolute values \(\sum \frac{1}{\sqrt{n+3}}\) diverges. This informs us that the given series is not absolutely convergent. Since the series passes the convergence test but not the absolute test, it converges conditionally. This is important to know because conditionally converging series are sensitive to rearrangements of terms which can affect the limit.

P-Test

The P-Test is a useful tool for determining the convergence of series in the form \(\sum \frac{1}{n^p}\). It compares the value of \(p\) to 1, which divides between divergence and convergence.- If \(p > 1\), the series converges.- If \(p \leq 1\), the series diverges.For our series' absolute values, \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+3}}\), the exponent looks like \(p = \frac{1}{2}\) because \(\frac{1}{\sqrt{n+3}} = \frac{1}{(n+3)^{1/2}}\). Since \(p = \frac{1}{2} < 1\), the series diverges by the P-Test. Remember, the P-Test provides a quick judgment on convergence and divergence, especially useful for series that resemble a simple power decay.