Problem 17 - Advanced Calculus

Question

Let \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots 2 n}\) (a) Prove that \(\left\\{a_{n}\right\\}\) is convergent. (b) Can you determine lim \(_{n \rightarrow \infty} a_{n} ?\)

Step-by-Step Solution

Answer

\((a)\) Yes, the sequence \(a_{n}\) is convergent. \((b)\) The limit as \(n \rightarrow \infty\) of \(a_{n}\) is 0.

Step 1: Analyze the Sequence

Given sequence \(a_{n}=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots 2n}\). We can rewrite each term in the sequence as \(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\), which makes our sequence look a little simpler.

Step 2: Prove Convergence

To prove that a sequence is convergent, we need to show that it is bounded and monotonic. Now, each term in our sequence \(a_{n+1}\) and \(a_{n}\) can be expressed as \(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} \cdot \frac{2n+1}{2n+2}\) and \(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n}\) respectively. Notice that \(a_{n+1}\) is produced from \(a_{n}\) by multiplying by \(\frac{2n+1}{2n+2}\), which is clearly less than 1. Thus, the sequence is decreasing and bounded below by zero. Hence, the sequence is convergent.

Step 3: Determine the Limit

With it established that the sequence is convergent, now let's find \(L = \lim_{n \rightarrow \infty} a_{n}\). From the previous step we know, \(a_{n+1} = a_{n} \cdot \frac{2n+1}{2n+2}\). In the limit as \(n \rightarrow \infty\), we can say \(a_{n} = L\). This implies \(L = L \cdot \frac{2n+1}{2n+2}\). On simplifying this, we get \(L = 0\). Thus, the limit of the sequence as \(n \rightarrow \infty\) is 0.

Key Concepts

Limit of a Sequence

Understanding the limit of a sequence is fundamental in calculus and analysis. A sequence is simply a list of numbers ordered in a specific way, typically expressed as \(a_1, a_2, a_3, \ldots\). The limit of a sequence \(\{a_n\}\) is the value that the sequence approaches as \(n\) becomes very large, or goes to infinity. To find the limit, we explore what happens to the terms of the sequence as \(n\) increases without bound. If the terms get closer and closer to a specific value, we say that the sequence converges to that value (the limit). Not all sequences have a limit; some might oscillate indefinitely or increase without bound. When a sequence does have a limit, it provides us with powerful insights into the behavior of the sequence as a whole. For the sequence given in the exercise, we showed that its limit is 0 as \(n\) approaches infinity, meaning the sequence gets arbitrarily close to 0 the further along it progresses.

Monotonic Sequence

A monotonic sequence is one that either never increases or never decreases. More clearly:

• If the sequence is never decreasing and always increasing, we call it non-decreasing.
• If it is never increasing and always decreasing, we call it non-increasing.

In our exercise, we considered the sequence \(a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n}\), and determined it was decreasing because each term \(a_{n+1}\) can be expressed as \(a_n\) times a fraction less than one. This consistent decrease classifies it as a monotonic sequence. Monotonic sequences are significant because if they are also bounded, they must have a limit. This property helps identify convergence without directly calculating the limit, providing a convenient tool in analysis.

Bounded Sequence

A bounded sequence is one where all its terms stay within fixed limits, known as bounds. Specifically, a sequence is bounded above if there is a number greater than or equal to every term in the sequence. Similarly, it is bounded below if there is a number less than or equal to every term in the sequence. If a sequence is both bounded above and bounded below, we say it is simply bounded. The sequence from our exercise is bounded below by 0, as every term is a product of fractions, ensuring the sequence does not dip below zero. The significance of a bounded sequence becomes clear when analyzing convergence. A sequence that is both bounded and monotonic is guaranteed to converge, which is why proving these two properties helped demonstrate that our sequence has a limit. Understanding whether a sequence is bounded is thus crucial for predicting its long-term behavior.