Question

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2^{\ln n}}$$

Short answer

The series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{2^{\ln n}}\) converges by the Alternating Series Test.

Step-by-step solution

- Check for zero using Preliminary Test

The preliminary test checks whether the terms of the series approach zero as n approaches infinity. Evaluate the limit: \ \( \lim_{{n \to \infty}} \frac{(-1)^n}{2^{\ln n}} \). The denominator \(2^{\ln n}\) grows very large because \(\ln n\) increases slowly but surely. Therefore, the entire fraction approaches zero: \ \( \lim_{{n \to \infty}} \frac{(-1)^n}{2^{\ln n}} = 0 \). The series passes the preliminary test.

- Choose the Alternating Series Test

Since the series has the form \( \sum (-1)^n a_n \) where \(a_n = \frac{1}{2^{\ln n}}\), the Alternating Series Test (Leibniz's Test) can be used. The test requires that: \ 1. The terms \(a_n\) are positive. \ 2. \( a_n \) is monotonically decreasing. \ 3. \( \lim_{{n \to \infty}} a_n = 0 \).

- Check if \(a_n = \frac{1}{2^{\ln n}}\) is Positive

Confirm that all terms \(a_n = \frac{1}{2^{\ln n}}\) are positive. Since \(2^{\ln n}\) is always positive for all \(n \in \mathbb{N}\), \(a_n\) will also be positive. This condition is satisfied.

- Verify Monotonic Decreasing Condition

Demonstrate that \(a_n = \frac{1}{2^{\ln n}}\) is decreasing. Compare terms \(a_n\) and \(a_{n+1}\): \ \( a_n = \frac{1}{2^{\ln n}} \) \ \( a_{n+1} = \frac{1}{2^{\ln (n+1)}} \). Notice \(\ln (n+1) > \ln n\), thus \(2^{\ln (n+1)} > 2^{\ln n}\), so \( \frac{1}{2^{\ln (n+1)}} < \frac{1}{2^{\ln n}} \), confirming \(a_n\) is decreasing.

- Confirm the Limit Condition

Validate \( \lim_{{n \to \infty}} \frac{1}{2^{\ln n}} = 0 \), as previously discussed in Step 1. \ Since \(2^{\ln n}\) grows without bound, the limit of \( \frac{1}{2^{\ln n}} \) as \(n\) approaches infinity is indeed 0.

- Conclude with Alternating Series Test Result

All conditions of the Alternating Series Test are satisfied: \ 1. \(a_n\) is positive. \ 2. \(a_n\) is decreasing. \ 3. \( \lim_{{n \to \infty}} a_n = 0 \). \ Therefore, the series \(\sum_{n=1}^{\infty} \frac{(-1)^n}{2^{\ln n}}\) converges by the Alternating Series Test.

Key concepts

Preliminary Test

The preliminary test is the initial step in checking if a series converges. It checks whether the terms of the series approach zero as the index goes to infinity. To perform the preliminary test, you calculate the limit of the sequence's general term. This step helps in quickly identifying divergent series. If the limit of the terms does not approach zero, the series is divergent, and there's no need to perform further tests.

For example, in our problem, we have the series \(\frac{(-1)^n}{2^{\ln n}}\). The preliminary test involves calculating:
\( \boxed{ \lim_{{n \to \infty}} \frac{{(-1)^n}}{2^{\ln n}} } \).

Since the denominator \(2^{\ln n}\) grows very large and the numerator alternates between -1 and 1, this fraction approaches zero. Hence, the series passes the preliminary test.

Alternating Series Test

The Alternating Series Test, also known as Leibniz's Test, is specifically used for series where the terms alternate in sign, such as those of the form \(\boxed{ \sum (-1)^n a_n}\). The test requires three conditions to be met:

• \(a_n\) must be positive.
• \(a_n\) must be monotonically decreasing.
• \( \lim_{{n \to \infty}} a_n = 0\).

Let's see how this is applied to the series \(\frac{(-1)^n}{2^{\ln n}}\). First, determine if \(a_n = \frac{1}{2^{\ln n}}\) is positive. Since \(2^{\ln n}\) is always positive, \(a_n\) is positive. Next, check if \(a_n\) is monotonically decreasing (explained in detail below) and finally, confirm that \( \lim_{{n \to \infty}} a_n = 0\), which we already know from the preliminary test. Since all conditions are satisfied, the series converges by the Alternating Series Test.

Monotonically Decreasing Function

A function or sequence is said to be monotonically decreasing if every term is less than or equal to the previous term. Verifying this property is crucial in convergence tests like the Alternating Series Test. To check if \(a_n = \frac{1}{2^{\ln n}}\) is decreasing, we compare it to the next term \(a_{n+1} = \frac{1}{2^{\ln (n+1)}}\). Since \(\ln (n+1) > \ln n\), it follows that \(2^{\ln (n+1)} > 2^{\ln n}\), making \( \frac{1}{2^{\ln (n+1)}} < \frac{1}{2^{\ln n}} \), thus proving that \(a_n\) is decreasing.

This step ensures that the sequence meets one of the critical conditions needed for the Alternating Series Test. It reinforces that our series \( \frac{(-1)^n}{2^{\ln n}} \) converges, as the terms are both positive and monotonically decreasing.