Problem 26 - Thomas Calculus

Question

Determining Convergence or Divergence 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 \(\lim_{n\to\infty} b_n \neq 0\).

Step 1: Check for Alternating Series Test

The given series is \[ \sum_{n=1}^{\infty}(-1)^{n}\left(1-\frac{3}{n}\right)^{n}. \]This is an alternating series because of the factor \((-1)^n\). We can apply the Alternating Series Test, which states that if \(b_n > 0\), \(b_n\) is decreasing, and \(\lim_{n\to\infty} b_n = 0\), then the series converges.

Step 2: Define Sequence Terms

Identify sequence terms:\[ b_n = \left(1-\frac{3}{n}\right)^{n}. \]

Step 3: Evaluate Limit of Sequence Terms

Let's find \(\lim_{n\to\infty} b_n\):\[ b_n = \left(1-\frac{3}{n}\right)^{n}. \]Using the limit property \(\lim_{n \to \infty} \left(1 - \frac{a}{n}\right)^n = e^{-a}\) for a constant \(a\), we have:\[ \lim_{n\to\infty} \left(1-\frac{3}{n}\right)^{n} = e^{-3}. \]

Step 4: Conclusion on Alternating Series Test

The Alternating Series Test requires \(\lim_{n\to \infty} b_n = 0\), but we found \(\lim_{n\to\infty} b_n = e^{-3} eq 0\). Therefore, the Alternating Series Test concludes that the series does not converge.

Step 5: Final Answer

Since the condition \(\lim_{n\to\infty} b_n = 0\) is not met, the series must diverge.

Key Concepts

Alternating Series Test

The Alternating Series Test is a handy tool when dealing with series that change sign with each term. These series are known as alternating series.
The given series, \[ \sum_{n=1}^{\infty} (-1)^{n} \left(1 - \frac{3}{n}\right)^{n} \]contains the factor \((-1)^n\), which signals that this series alternates. This feature allows us to use the Alternating Series Test.
To apply this test, we look for the non-alternating part of the series, \(b_n = \left(1 - \frac{3}{n}\right)^{n} \), and analyze it based on three conditions:

• Each term, \(b_n\), should be positive.
• \(b_n\) should be decreasing as \(n\) becomes larger.
• The limit of \(b_n\) as \(n\) approaches infinity should be zero.

When these conditions hold, the series is convergent according to the test. However, if any condition is not satisfied, the series diverges. In our example, we find that the limit isn't zero, leading us to conclude that the series diverges.

Sequence Limits

Sequence limits are crucial for understanding the behavior of series terms as they grow very large.
To determine the limit of the sequence terms \(b_n = \left(1 - \frac{3}{n}\right)^{n}\), we leverage a well-known limit property: \[ \lim_{n \to \infty} \left(1 - \frac{a}{n}\right)^n = e^{-a} \]where \(a\) is a constant.
Applying this property to our sequence, with \(a = 3\), gives:\[\lim_{n\to\infty} \left(1 - \frac{3}{n}\right)^{n} = e^{-3}.\]Here, unlike in simple cases where the sequence limit is zero, our result is a positive number, \(e^{-3}\).This result directly affects the alternating series test, as it invalidates the test's convergence condition which requires the limit to be zero.Understanding sequence limits helps predict whether series will add up to a finite sum or grow indefinitely.

Infinite Series

An infinite series is the sum of an endless succession of terms. Understanding whether such a series converges or diverges is vital. It determines if adding infinitely many terms results in a finite number or keeps growing.The series \[ \sum_{n=1}^{\infty} (-1)^{n} \left(1 - \frac{3}{n}\right)^{n} \]continues indefinitely. Each term is influenced by its sequence form, the alternating sign, and the formula \(\left(1-\frac{3}{n}\right)^{n}\).
Analyzing these terms gives us insight into their contribution to the whole series. In this example, applying various tests like the Alternating Series Test helps to conclude that each term's diminishing pattern does not satisfy convergence. Thus, no finite sum can fully encompass the output of all the terms, resulting in the series diverging.By understanding infinite series and convergence criteria, it becomes easier to grasp whether these series depict real-world situations or result in mathematical inconsistencies.