Problem 45 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n} \operatorname{sech} n $$

Step-by-Step Solution

Answer

The series converges absolutely by the Alternating Series Test and Comparison Test.

Step 1: Understanding the Series Type

The given series is \( \sum_{n=1}^{\infty}(-1)^{n} \text{sech} \, n \). This series is an alternating series as it involves the term \((-1)^{n}\). Alternating series can be checked for convergence using the Alternating Series Test.

Step 2: Apply the Alternating Series Test

To apply the Alternating Series Test, we need to check two conditions: 1) The absolute value of the terms \( \text{sech} \, n \) should be decreasing. 2) The terms \( \text{sech} \, n \) should approach zero as \( n \to \infty \). Since \( \text{sech} \, n = \frac{2}{e^n + e^{-n}} \), both conditions are satisfied because \( \text{sech} \, n \) decreases and \( \text{sech} \, n \rightarrow 0 \) as \( n \) increases.

Step 3: Consider Absolute Convergence

For absolute convergence, we consider the series \( \sum_{n=1}^{\infty} \left| \text{sech} \, n \right| = \sum_{n=1}^{\infty} \text{sech} \, n \). Since \( \text{sech} \, n \) is always positive, the series is the same as the original without the alternating sign. We compare this to the series \( \sum_{n=1}^{\infty} e^{-n} \), a geometric series with ratio \(r = \frac{1}{e} < 1\). Since the geometric series converges, the series \( \sum_{n=1}^{\infty} \text{sech} \, n \) converges.

Step 4: Conclusion on Convergence

Since the series \( \sum_{n=1}^{\infty} \text{sech} \, n \) converges, the original series \( \sum_{n=1}^{\infty} (-1)^n \text{sech} \, n \) converges absolutely. Absolute convergence implies convergence, so we conclude that the given series converges absolutely.

Key Concepts

Alternating Series Test

In the world of infinite series, not all series behave the same way. Some series alternate between positive and negative terms, and these are called alternating series. The series \( \sum_{n=1}^{\infty} (-1)^{n} \text{sech} \, n \) is an example of this because it switches signs due to

• the \((-1)^n\) factor, which causes the terms to alternate between positive and negative values.

To determine if an alternating series converges, we use the Alternating Series Test. This test provides two key conditions that must be met:

• The absolute value of the terms \(\text{sech} \, n\) should decrease steadily.
• The terms \(\text{sech} \, n\) should approach zero as \(n\) goes to infinity.

When both conditions hold true, the series converges. In our case, since \(\text{sech} \, n = \frac{2}{e^n + e^{-n}}\), it is clear that:

• \(\text{sech} \, n\) decreases because the exponential term \(e^n\) grows rapidly, making the denominator bigger as \(n\) increases.
• \(\text{sech} \, n\) approaches zero as \(n\) approaches infinity, since the denominator becomes huge in comparison to the numerator.

Thus, this alternating series converges according to the Alternating Series Test.

Absolute Convergence

Absolute convergence is a stronger form of convergence where a series not only converges normally but also converges when we take the absolute value of its terms. For the series \( \sum_{n=1}^{\infty} (-1)^{n} \text{sech} \, n \), we consider the series without the alternating sign: \( \sum_{n=1}^{\infty} \left| \text{sech} \, n \right| = \sum_{n=1}^{\infty} \text{sech} \, n \).

• This leads us to examine if the positive series still converges.

When a series converges absolutely, it implies the series also converges in the usual sense but is more robust:

• If a series converges absolutely, it is guaranteed to converge even without considering alternation.

By comparing \( \sum_{n=1}^{\infty} \text{sech} \, n \) to the geometric series \( \sum_{n=1}^{\infty} e^{-n} \), with a ratio \( r = \frac{1}{e} < 1 \), we establish:

• Convergence of \(\sum_{n=1}^{\infty} \text{sech} \, n\) since it follows a pattern similar to the geometric series with terms decreasing and approaching zero.

Therefore, because \( \sum_{n=1}^{\infty} \text{sech} \, n \) converges like the geometric series, the original series \( \sum_{n=1}^{\infty} (-1)^{n} \text{sech} \, n \) is absolutely convergent. Hence, it converges absolutely, confirming the series is well-behaved with respect to both convergence tests.

Geometric Series

A geometric series is a type of series where each term is a constant multiple of the term before it. It can be written as \( \sum_{n=0}^{\infty} ar^n \) where \(a\) is the first term and \(r\) is the common ratio.

• Geometric series have a unique convergence rule: they converge if the absolute value of the common ratio \(|r|\) is less than 1.

One essential geometric series is \( \sum_{n=1}^{\infty} e^{-n} \), which has:

• Terms forming a geometric sequence with \( r = \frac{1}{e} \).
• Because \( \frac{1}{e} < 1 \), this series converges.

This convergence property helps us in proving the convergence of other series. In our case, it confirms that \( \sum_{n=1}^{\infty} \text{sech} \, n \) converges by comparison:

• Each \( \text{sech} \, n\) term behaves similarly to \(e^{-n}\), exhibiting a diminishing nature.

Thus, by the properties of geometric series, it supports our conclusion that the original alternating series we've studied converges absolutely. Utilizing these features aids in verifying the convergence and absolute convergence of more complex series.