Problem 7 - Thomas Calculus

Question

Determine if the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{n^{2}} $$

Step-by-Step Solution

Answer

The series diverges because \( \lim_{n \to \infty} \frac{2^{n}}{n^{2}} \neq 0 \).

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{n^{2}} \). This is an alternating series, where the terms alternate in sign.

Step 2: Define the Sequence for the Alternating Series Test

For the alternating series test, the sequence without the alternating factor is \( a_n = \frac{2^{n}}{n^{2}} \). We need to check if this sequence satisfies the conditions for the test.

Step 3: Check Condition 1: Limit of the Sequence

Check if \( \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{2^{n}}{n^{2}} = 0 \). Since exponential growth \(2^n\) dominates \(n^2\), the limit does not tend to zero. Thus, \( \lim_{n \to \infty} a_n eq 0 \).

Step 4: Conclusion Based on the Limit

Since \( \lim_{n \to \infty} a_n eq 0 \), the alternating series test requires this condition to be zero for convergence, which is not satisfied. As a result, the series does not converge.

Key Concepts

Convergence of Series

In mathematics, establishing the convergence of a series is essential to understand the behavior of its sum as the number of terms tends to infinity. When a series converges, it means that adding up all the terms in the series leads to a finite value. Alternating series are a specific kind of series where the signs of the terms switch between positive and negative.One of the primary tools to judge convergence in an alternating series is the Alternating Series Test. This test states that an alternating series \[ \sum_{n=1}^{\infty} (-1)^{n+1} a_n \]converges if:

• The absolute value of the terms \(a_n\) is decreasing, i.e., \(a_n > a_{n+1}\) for all \(n\).
• The limit of \(a_n\) as \(n\) approaches infinity is zero, \( \lim_{n \to \infty} a_n = 0 \).

In the provided series, \(a_n = \frac{2^n}{n^2}\), and the test fails because the terms do not decrease to zero. This indicates divergence, which we'll discuss further in the next section.

Divergence of Series

When a series diverges, it means that the sum of its infinite terms grows without bound, or oscillates without settling at a particular value. For the alternating series given in the problem, the key reason for divergence lies in the balance between the exponential growth of \(2^n\) and the polynomial growth of \(n^2\).The Alternating Series Test, which would help confirm convergence, requires two conditions: The terms should get smaller as the series progresses, and the limit of the terms should approach zero. In the sequence \(a_n = \frac{2^n}{n^2}\), neither of these conditions is satisfied.

• First, instead of decreasing, \(2^n\) grows much faster than \(n^2\), meaning that \(a_n\) does not approach zero as \( n \) becomes very large.
• Second, since the limit of \(a_n\) is not zero, this essential condition for convergence fails.

Thus, by the Alternating Series Test, this series does not converge, reaffirming its divergence.

Exponential Growth vs Polynomial Growth

Understanding the distinction between exponential and polynomial growth is pivotal in series analysis, especially in determining convergence or divergence. In the provided series, the sequence \(a_n = \frac{2^n}{n^2}\) encapsulates this contrast neatly.Exponential growth describes processes like \(2^n\), where the base of the exponential is a constant raised to a power that increases linearly. This type of growth outpaces any polynomial growth as \(n\) approaches infinity.

• Polynomial growth, like \(n^2\), increases at a slower rate, such as squaring or cubing terms as they grow linearly.
• In most contexts involving series, exponential terms will dominate polynomial terms given enough large values of \(n\).

For our sequence, \(2^n\) grows far too quickly compared to the slowing polynomial \(n^2\), leading \(a_n = \frac{2^n}{n^2}\) to increase rather than decrease to zero. This fact supports the conclusion of divergence drawn from the Alternating Series Test.