Problem 38 - Essential Calculus

Question

Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{n !}\)

Step-by-Step Solution

Answer

The sequence \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{n !}\) is divergent.

Step 1: Express the next term

Express \(a_{n+1}=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n + 1)}{(n+1) !}\)

Step 2: Compute ratio

Compute the ratio \(R= \frac{a_{n+1}}{a_n} ~=~ \frac{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n + 1)}{(n+1) !}}{\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{n !}}\). Simplify the ratio to \(R=\frac{2n+1}{n+1}\)

Step 3: Apply Ratio Test

For the Ratio Test, if the absolute value of \(R\) is less than 1, the series is absolutely convergent. If it is more than 1, it is divergent, and if it is equal to 1, the test is inconclusive. Calculate the limit as \(n\) tends to infinity. The limit here is \( \lim_{n \to \infty} \frac{2n+1}{n+1} = 2 \)

Step 4: Conclusion

Since the limit of the absolute value of the ratio is greater than 1, it can be concluded that the sequence is divergent

Key Concepts

Ratio Test

The Ratio Test is a handy tool for determining the convergence of infinite series or sequences. Consider a sequence denoted by \(a_n\). The test involves expressing the next term, \(a_{n+1}\), and forming the ratio \(R = \frac{a_{n+1}}{a_n}\). This ratio is then simplified, and its limit as \(n\) approaches infinity is calculated.

Here's a quick summary of what to do with this limit:

• If \(|R| < 1\), the sequence converges absolutely.
• If \(|R| > 1\), the sequence diverges.
• If \(|R| = 1\), the test is inconclusive, and another method must be used.

In our exercise, after calculating \(R = \frac{2n+1}{n+1}\), we found that as \(n\) tends towards infinity, the limit is 2. Since this limit is greater than 1, the sequence diverges.

Divergence of Sequence

A sequence diverges when its terms do not approach a fixed value as the sequence progresses. Essentially, the sequence does not settle into a single place. Instead, the terms can keep increasing, decreasing, or oscillating without converging to a specific number.

In our original exercise, we applied the ratio test and calculated a limit \(R\) that was greater than 1. This outcome indicates divergence.

To visualize this, think about it like walking on a path without ever stopping or finding a set destination. The sequence keeps moving in that way without ever resting at one precise point.

Limit of a Sequence

When a sequence converges, its terms get closer and closer to a particular value. This is where the idea of a limit comes into play. The limit of a sequence \(a_n\) as \(n\), the term number, approaches infinity, is the specific number that \(a_n\) gets ever closer to. Such a number is called the sequence's limit and is denoted as \(\lim_{n \to \infty} a_n = L\), where \(L\) would be the limit.

For example, consider a sequence like \(a_n = \frac{1}{n}\). Here, as \(n\) grows larger, \(a_n\) approaches 0 - making 0 the limit of the sequence. However, if a sequence is divergent, it means there is no such fixed value that \(a_n\) approaches.

In the original problem, since the sequence diverged, there isn't a limit. The challenge in sequence analysis often lies in correctly determining whether a limit exists by using convergence tests and mathematical reasoning.