Problem 36 - Essential Calculus

Question

Determine whether the series converges conditionally or absolutely, or diverges. $$ \sum_{n=0}^{\infty}(-1)^{n} e^{-n^{2}} $$

Step-by-Step Solution

Answer

The series \(\sum_{n=0}^{\infty}(-1)^{n} e^{-n^{2}}\) is absolutely convergent.

Step 1: Check Convergence Using the Alternating Series Test

Alternating series test checks two conditions: (\(a_{n+1} ≤ a_{n}\) for all \(n\)) and (\(lim_{n→∞} a_{n} = 0\)). Here, \(a_{n} = e^{-n^{2}}\). As \(n\) increases, since the exponent of \(e\) is being squared and negated, the sequence \(a_{n}\) is decreasing to 0. Therefore, the series \(\sum_{n=0}^{\infty}(-1)^{n} e^{-n^{2}}\) converges conditionally.

Step 2: Check Absolute Convergence

Absolute convergence requires that the absolute value of the series \(\sum_{n=0}^{\infty} |(-1)^{n} e^{-n^{2}}|\) or \(\sum_{n=0}^{\infty} e^{-n^{2}}\) also converges. For comparison, consider the series \(\sum_{n=0}^{\infty} e^{-n}\). Note that for all \(n\), \(e^{-n} \geq e^{-n^{2}}\), because raising any positive number (in this case \(e^{-n}\)) to a power (squared here) only makes it smaller as the power grows. The series \(\sum_{n=0}^{\infty} e^{-n}\) is a convergent geometric series with ratio lesser than 1 (\(e^{-1}\)). By the comparison test, the series \(\sum_{n=0}^{\infty} e^{-n^{2}}\) also converges.

Key Concepts

Alternating Series Test

The Alternating Series Test is a critical technique for determining the convergence of infinite series whose terms alternate in sign. This method checks whether the absolute value of sequence elements decreases monotonically and approaches zero as the index grows. Specifically, let's consider a series \( \sum_{n=1}^\infty (-1)^{n} a_n \), where the sign alternates due to \( (-1)^{n} \). For the test to confirm convergence, two conditions must be met: each term \( a_n \) must be less than or equal to the previous term \( a_{n-1} \), and \( \lim_{n\to\infty} a_n = 0 \).

If these conditions are satisfied, the series converges conditionally. This means the series converges due to the alternation of signs, despite individual terms not necessarily forming a convergent series themselves if considered without signs.

Absolute Convergence

Absolute convergence is a stronger form of convergence for an infinite series. A series \( \sum_{n=1}^\infty a_n \) is said to be absolutely convergent if the series of absolute values \( \sum_{n=1}^\infty |a_n| \) also converges. When a series converges absolutely, it will also converge conditionally. However, the reverse is not always true: conditional convergence does not guarantee absolute convergence.

Checking for absolute convergence often involves using other tests for positive series, like the Comparison Test or the Ratio Test. This is because when we take the absolute value of the terms in the original series, we obtain a new series with all positive terms, which simplifies the analysis.

Geometric Series

A geometric series is an infinite series of the form \( \sum_{n=0}^\infty ar^n \), where \( a \) represents the first term, and \( r \) is the constant ratio between successive terms. A geometric series will converge if the absolute value of the ratio is less than one \( (|r| < 1) \), and its sum can be found using the formula \( S = \frac{a}{1 - r} \) where \( S \) is the sum of the series. If \( |r| \geq 1 \), the series will diverge.

When analyzing convergence of series that resemble a geometric series, identifying the ratio \( r \) and verifying \( |r| < 1 \) is crucial. This provides a straightforward check for the convergence of many series with terms involving powers of \( n \).

Comparison Test

The Comparison Test is an invaluable tool for determining the convergence or divergence of an infinite series by comparing it to a second series whose convergence behavior is already known. When we have a series \( \sum_{n=1}^\infty a_n \) where all \( a_n \) are positive, and a second series \( \sum_{n=1}^\infty b_n \) with known convergence or divergence, two outcomes are possible.

• If \( \sum_{n=1}^\infty b_n \) converges, and \( a_n \leq b_n \) for all \( n \), then \( \sum_{n=1}^\infty a_n \) also converges.
• If \( \sum_{n=1}^\infty b_n \) diverges, and \( a_n \geq b_n \) for all \( n \), then \( \sum_{n=1}^\infty a_n \) also diverges.

This test is particularly handy when the series in question is complex, but can be bounded by simpler expressions for which we can easily determine convergence properties.