Question

An alternating series \(\sum_{n=i}^{\infty} a_{n}\) is given. (a) Determine if the series converges or diverges. (b) Determine if \(\sum_{n=0}^{\infty}\left|a_{n}\right|\) converges or diverges. (c) If \(\sum_{n=0}^{\infty} a_{n}\) converges, determine if the convergence is conditional or absolute. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}$$

Short answer

(a) The series converges. (b) \(\sum |a_n|\) diverges. (c) The convergence is conditional.

Step-by-step solution

Identify the series

The given series is an alternating series:\[ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}. \]It can be identified by the presence of \((-1)^n\), which alternates the sign of each term.

Apply the Alternating Series Test

The Alternating Series Test states that the series \(\sum (-1)^n a_n\) converges if the following conditions are satisfied:1. \(a_n > 0\) for all \(n\).2. \(a_n\) is a decreasing sequence.3. \(\lim_{n \to \infty} a_n = 0\).For our series, \(a_n = \frac{1}{\sqrt{n}}\):- Clearly, \(a_n > 0\) for all \(n\) since the square root of a positive number is positive.- \(a_n\) is decreasing because \(\frac{1}{\sqrt{n}}\) decreases as \(n\) increases.- \(\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0\).Hence, by the Alternating Series Test, the series converges.

Check Absolute Convergence

We need to check the convergence of \(\sum_{n=0}^{\infty}\left|a_{n}\right|\), which is:\[ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}.\]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 \(\frac{1}{2} < 1\), this series diverges.

Determine Type of Convergence

Since the original series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n}}\) converges, but the absolute series \(\sum_{n=0}^{\infty}\left|a_{n}\right| = \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) diverges, the series converges conditionally.

Key concepts

Conditional Convergence

Conditional convergence occurs in series where the series converges, but the series of the absolute values of its terms does not converge. In simpler terms, a conditionally convergent series relies on the alternating nature of its terms to approach a limit.
Let's explore this with our given series:

• The series is \[ \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}} \].
• By passing the Alternating Series Test, it is established that the series itself converges.
• However, when taking the absolute value of each term and exploring \[ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \], this series diverges as it is a p-series with \( p = \frac{1}{2} \), which is less than 1.

The convergence of the original series without absolute values tells us it converges conditionally due to its alternating nature, even though the absolute values diverge.

Absolute Convergence

In contrast to conditional convergence, absolute convergence happens when a series converges even when you replace each term with its absolute value. It is a stronger condition than conditional convergence.
For absolute convergence:

• Take the given alternating series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}} \).
• Convert it into the absolute value series: \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \).
• This absolute series behaves like a p-series with \( p = \frac{1}{2} \), and we know p-series \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) converges only if \( p > 1 \).
• Therefore, since \( p = \frac{1}{2} \), the absolute series diverges.

Since the series of absolute values does not converge, the original series cannot be absolutely convergent. This distinction is essential for understanding how different series types behave.

P-Series Convergence

P-series convergence focuses on series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Understanding the behavior of p-series can help with evaluating convergence.
For a p-series:

• If \( p > 1 \), the series converges.
• If \( 0 < p \leq 1 \), the series diverges.

In our case:

• When examining the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \), it mirrors a p-series with \( p = \frac{1}{2} \).
• Since \( \frac{1}{2} \) is less than 1, the series prediction is for divergence.

Understanding p-series convergence is vital since it provides quick insights into the behavior of series based solely on the value of p. This knowledge simplifies the process of evaluating series like the one in our example.