Question

Determine whether the series \(\sum_{n=1}^{\infty}(-1)^{n+1} n / 2^{n}\) converges or diverges.

Short answer

The series converges by the alternating series test.

Step-by-step solution

Identify the Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} \). It alternates signs, as indicated by the \((-1)^{n+1}\) term. This is an alternating series.

Use the Alternating Series Test

The alternating series test states that a series \( \sum (-1)^{n} a_{n} \) converges if: 1. \( a_{n} > a_{n+1} \) for all \( n \) (the sequence is decreasing), and 2. \( \lim_{n \to \infty} a_{n} = 0 \). Here, \( a_{n} = \frac{n}{2^{n}} \), which we need to analyze.

Check if the Sequence is Decreasing

The sequence \( a_{n} = \frac{n}{2^{n}} \) should be checked for whether it is decreasing for all \( n \). As \( n \) increases, although \( n \) increases, the term \( 2^{n} \) grows exponentially faster. Thus, \( \frac{n}{2^{n}} > \frac{n+1}{2^{n+1}} \) for all positive \( n \), confirming the sequence is decreasing.

Check if the Limit is Zero

Evaluate \( \lim_{n \to \infty} \frac{n}{2^{n}} \) using L'Hôpital's Rule. As \( n \to \infty \), the exponential in the denominator dominates, leading to:\[ \lim_{n \to \infty} \frac{n}{2^{n}} = 0. \]Therefore, the sequence \( a_{n} \) satisfies the second condition of the alternating series test.

Conclusion on Convergence

Both conditions of the alternating series test hold: the sequence \( \frac{n}{2^{n}} \) is decreasing, and its limit as \( n \to \infty \) is 0. Thus, by the alternating series test, the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n}{2^{n}} \) converges.

Key concepts

Series Convergence

When discussing series, the term "convergence" refers to whether the sum of the infinite sequence of terms approaches a finite number. There are different types of series, and each type has distinct methods to test for convergence.

• An infinite series converges if the sequence of its partial sums has a limit, which is a real number.
• If a series does not converge, it is said to diverge, meaning the partial sums do not settle to a finite number.

For the series in this exercise, it's crucial to check the behavior of the series as the number of terms increases infinitely. Here, we use the Alternating Series Test to investigate the convergence. It's a perfect fit for examining series where the terms alternate in sign.

Alternating Series Test

The Alternating Series Test is a powerful tool used to determine the convergence of series where terms alternate in sign. This particular test applies to series of the form \( \sum (-1)^{n} a_{n} \), where the term \((-1)^{n}\) makes the sequence's signs alternate from positive to negative.
To utilize this test, two main conditions must be satisfied:

• The sequence of absolute values \(a_{n}\) must be decreasing; specifically, \(a_{n} \ge a_{n+1}\) for all \(n\).
• The limit of \(a_{n}\) as \(n\) approaches infinity must be zero: \(\lim_{n \to \infty} a_{n} = 0\).

If both conditions are met, the series converges. In our original problem, the sequence \( \frac{n}{2^{n}} \) is indeed decreasing, and its limit is zero, ensuring the series converges.

Limits

Limits are a fundamental concept when analyzing series convergence. When proving whether a series converges using tests like the Alternating Series Test, we often need to find the limit of a sequence.
In the current exercise, assess the limit of the sequence \( \frac{n}{2^{n}} \) as \(n\) approaches infinity. Even though \(n\) increases, the exponential term \(2^{n}\) in the denominator grows much faster.
Thus, \( \frac{n}{2^{n}} \) tends to zero as \(n\) approaches infinity. This conclusion is essential in confirming the series converges by the Alternating Series Test. By establishing a limit of zero, the series satisfies a critical condition for convergence.

L'Hôpital's Rule

L'Hôpital's Rule is a technique in calculus used to find limits of indeterminate forms. If you encounter limits like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule allows for differentiation of the numerator and denominator.
This rule is particularly handy when evaluating the limit of sequences where direct substitution isn't straightforward. For example, in the provided problem, the limit of \(\frac{n}{2^{n}}\) as \(n\) approaches infinity can be hard to calculate directly.
By differentiating the numerator and denominator individually, you can simplify the expression to find \( \lim_{n \to \infty} \frac{n}{2^{n}} = 0 \). Utilizing L'Hôpital's Rule in such cases provides a clearer path to establishing the convergence of the series.