Problem 25 - Thomas Calculus

Question

In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty}(-1)^{n}\left(1-\frac{3}{n}\right)^{n}$$

Step-by-Step Solution

Answer

The series diverges because the terms do not tend to zero.

Step 1: Examine the series expression

The series in question is \( \sum_{n=1}^{\infty}(-1)^{n}\left(1-\frac{3}{n}\right)^{n} \). It is an alternating series due to the \((-1)^n\) factor, which represents changes in sign with each term.

Step 2: Simplify and examine the term \(a_n\)

Consider the term inside the series \( \left(1-\frac{3}{n}\right)^{n} \). As \( n \to \infty \), this approaches \( e^{-3} \) because it is of the form of \( \left(1 + \frac{x}{n}\right)^n \rightarrow e^x \). Thus, the sequence \( \left(1 - \frac{3}{n}\right)^n \) approaches \( e^{-3} \) as \( n \to \infty \).

Step 3: Apply the Alternating Series Test

For the alternating series \( \sum (-1)^n a_n \), where \( a_n = \left(1-\frac{3}{n}\right)^{n} \), to converge the following must hold for \( a_n \): (i) \( a_n \to 0 \) as \( n \to \infty \), and (ii) \( a_{n+1} \le a_n \) for all \( n \). Here, \( a_n \) does not approach zero; it approaches \( e^{-3} eq 0 \).

Step 4: Conclude the divergence due to failing conditions

Since the first condition of the Alternating Series Test (convergence requires \( a_n \to 0 \)) is not satisfied, the series \( \sum_{n=1}^{\infty}(-1)^{n}\left(1-\frac{3}{n}\right)^{n} \) diverges.

Key Concepts

Alternating Series Test

When examining if a series converges, the Alternating Series Test is a valuable method. This test applies to series whose terms alternate in sign, like the series given in the exercise involving \((-1)^n\), where the signs switch from positive to negative as \(n\) becomes odd or even.
The Alternating Series Test tells us that for an alternating series \(\sum (-1)^n a_n\) to converge, two conditions must be satisfied:

• The sequence \(a_n\) must converge to zero as \(n\) approaches infinity, which means \(a_n \to 0\).
• The terms \(a_{n+1}\) must not be larger than the previous terms \(a_n\) for each \(n\), which is stated as \(a_{n+1} \le a_n\).

In our exercise, the term \(a_n\) is \(\left(1-\frac{3}{n}\right)^{n}\). As \(n\) becomes very large, \(a_n\) approaches \(e^{-3}\) rather than zero, failing the first condition. This means the series does not meet the criteria of the Alternating Series Test for convergence.

Limits

Limits help us understand how a function behaves as it approaches a particular point, often infinity. In this exercise, we need to find the limit of the expression \(\left(1-\frac{3}{n}\right)^{n}\) as \(n\) approaches infinity.
We realize this expression is similar to the compound interest formula that approaches the limit \(e^x\) as \(n\) becomes very large. Specifically, this is a case where the expression \(\left(1+\frac{x}{n}\right)^n\) approaches \(e^x\) when evaluated at infinity, meaning our series term \(\left(1-\frac{3}{n}\right)^{n}\) indeed approaches \(e^{-3}\).
Understanding this limit is crucial for determining if \(a_n\) converges to zero or not, which it does not in this case.

Divergence

Divergence in series indicates that adding up an infinite number of terms does not result in a finite value. Instead, it suggests that the series increases without bound or oscillates, failing to settle to a particular value.
In this exercise, because \(a_n\) approaches \(e^{-3}\), which is not zero, the series inherently fails to meet the necessary condition under the Alternating Series Test for convergence. This means our series does not reach a fixed sum and is thus divergent.
Recognizing divergence is important as it reflects the nature of the series and guides us in understanding behaviors associated with infinite series. In this case, since one of the convergence criteria is not met, we can confidently conclude that the series diverges.