Problem 24 - Single Variable Calculus

Question

Determine whether the series converges or diverges. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt[n]{n}} $$

Step-by-Step Solution

Answer

The given alternating series converges, since the sequence of absolute values is decreasing and the limit of the sequence of absolute values approaches 0.

Step 1: Identify the sequence \(a_n\) of the series

We are given the alternating series: \[ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt[n]{n}} \] First, we identify the sequence \(a_n\) of the series as: \[ a_n = \frac{(-1)^{n-1}}{\sqrt[n]{n}} \]

Step 2: Evaluate the absolute value of the sequence

Next, we find the absolute value of the sequence \(a_n\): \[ |a_n| = \left|\frac{(-1)^{n-1}}{\sqrt[n]{n}}\right| = \frac{1}{\sqrt[n]{n}} \]

Step 3: Check if the sequence of absolute values is decreasing

We need to determine if the sequence of absolute values is decreasing: \[ |a_{n+1}| = \frac{1}{\sqrt[{n+1}]{n+1}} \le \frac{1}{\sqrt[n]{n}} = |a_n| \] Since the denominator is always increasing (larger exponent roots), we can ensure that |a_{n+1}| is less than |a_n|. Thus, the sequence of absolute values is decreasing.

Step 4: Evaluate the limit of the absolute value sequence

We need to find the limit of the sequence of absolute values as n approaches infinity: \[ \lim_{n \to \infty} |a_n| = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} \] Using the property that \(\lim_{n \to \infty} \sqrt[n]{n} = 1\), we get: \[ \lim_{n \to \infty} |a_n| = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = \frac{1}{1} = 0 \]

Step 5: Apply the Alternating Series Test

Since both conditions of the Alternating Series Test are met: 1. The sequence of absolute values is decreasing: \(|a_{n+1}| \le |a_n|\) 2. The limit of the sequence of absolute values approaches 0: \(\lim_{n \to \infty} |a_n| = 0\) Therefore, the given alternating series converges.

Key Concepts

Alternating Series Test

The Alternating Series Test is an effective method to determine the convergence of series that alternate in sign. For a given series \( \sum (-1)^{n-1}b_n \), where \( b_n \) are positive terms, two primary conditions must be satisfied to conclude convergence:

• The terms \( b_n \) must be monotonically decreasing for all \( n \ge N \), which means each subsequent term should be smaller than or equal to the previous one.
• The limit of \( b_n \) as \( n \) approaches infinity must equal zero: \( \lim_{{n \to \infty}} b_n = 0 \).

When both of these conditions are fulfilled, the alternating series converges. This test is especially helpful as it provides convergence information without requiring the full sum of the series to be calculated. As seen in the exercise, confirming both conditions for the sequence \( \frac{1}{\sqrt[n]{n}} \) allows us to assert that the series converges.

Sequence of Absolute Values

The sequence of absolute values in a series is crucial for applying tests like the Alternating Series Test. By examining \( |a_n| = \left| \frac{(-1)^{n-1}}{\sqrt[n]{n}} \right| = \frac{1}{\sqrt[n]{n}} \), we focus only on the series' magnitude without worrying about sign changes. Understanding the nature of this sequence aids in evaluating whether it is decreasing, a necessary condition for convergence in alternating series. Specifically:

• Calculate \( |a_{n+1}| \) and compare it to \( |a_n| \) to check for monotonic decrease.
• In our example, as \( n \) increases, \( \sqrt[n]{n} \) grows, leading to smaller fractions, hence a decreasing sequence.

Verifying these properties enables us to use convergence tests confidently, knowing we meet required conditions.

Limit of a Sequence

Finding the limit of a sequence is integral in assessing series convergence. For the sequence of interest \( |a_n| = \frac{1}{\sqrt[n]{n}} \), calculating its limit as \( n \to \infty \) helps confirm series behavior at infinity.The specific sequence \( \sqrt[n]{n} \) is pivotal here. It's understood that:

• As \( n \) grows, \( \sqrt[n]{n} \) approaches 1 because the incremental effect of taking the nth root essentially flattens out the sequence.
• The limiting behavior results in \( \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = \frac{1}{1} = 0 \).

Reaching a limit of zero satisfies a critical part of many convergence tests, including the Alternating Series Test, confirming as in our textbook solution that the series indeed converges.