Problem 99 - Essential Calculus

Question

Consider the sequence \(\left\\{a_{n}\right\\}\) where \(a_{1}=\sqrt{k}, a_{n+1}=\sqrt{k+a_{n}}\), and \(k>0\) (a) Show that \(\left\\{a_{n}\right\\}\) is increasing and bounded. (b) Prove that \(\lim _{n \rightarrow \infty} a_{n}\) exists. (c) Find \(\lim _{n \rightarrow \infty} a_{n}\).

Step-by-Step Solution

Answer

Part (a): the sequence \( \{a_{n}\} \) is increasing and bounded by 0 and \(\sqrt{2k}\).\nPart (b): By the Monotone Convergence Theorem, the sequence converges, which means \(\lim _{n \rightarrow \infty} a_{n}\) exists.\nPart (c): The limit of the sequence as \(n\) approaches infinity is \( \frac{1+ \sqrt{1+4k}}{2}\).

Step 1: Proving the sequence is increasing

Firstly, apply induction to show the sequence is increasing. We know that \(a_{1}= \sqrt{k}\) and \(a_{2}= \sqrt{k+ \sqrt{k}} > \sqrt{k}= a_{1}\). Thus, the inequality holds true for \(n=1\). Assume that \(a_{n} < a_{n+1}\), this is the induction hypothesis. We want to show that \(a_{n+1} < a_{n+2}\). Use the induction hypothesis to get \(a_{n+2}= \sqrt{k+a_{n+1}} > \sqrt{k + a_{n}}= a_{n+1}\). Thus, the sequence \(a_{n}\) is increasing.

Step 2: Proving the sequence is bounded

Next, show the sequence is bounded. Clearly, \(a_{1}= \sqrt{k} > 0\), and suppose \(a_{n}>0\), then \(a_{n+1}= \sqrt{k+ a_{n}} > 0\). Thus, \(a_{n} > 0\) for all \(n \geq 1\). Also, \(a_{1}= \sqrt{k}< \sqrt{2k}\), and suppose \(a_{n}< \sqrt{2k}\), then \(a_{n+1}= \sqrt{k+a_{n}} < \sqrt{2k}\), yielding \(a_{n}< \sqrt{2k}\) for all \(n \geq 1\). Therefore, the sequence \(a_{n}\) is bounded by 0 and \(\sqrt{2k}\).

Step 3: Prove that the limit exists

Since the sequence is increasing and bounded, by the Monotone Convergence Theorem, the sequence converges, implying that \(\lim _{n \rightarrow \infty} a_{n}\) exists.

Step 4: Find the limit

Let \(L = \lim _{n \rightarrow \infty} a_{n}\). Then, \(L = \lim_{n \rightarrow \infty} a_{n+1}= \lim_{n \rightarrow \infty} \sqrt{k+a_{n}} = \sqrt{k+L}\), squaring both sides and simplifying gives \(L= \frac{1+ \sqrt{1+4k}}{2}\)

Key Concepts

Bounded Sequences

A sequence is considered bounded if it stays within fixed limits as it progresses. Think of an invisible boundary line that the sequence never crosses. For the sequence \( \{a_n\} \), we showed it's bounded between 0 and \( \sqrt{2k} \).

Here's how it works:

• We start with \( a_1 = \sqrt{k} \), clearly greater than 0.
• For each term, \( a_{n+1} = \sqrt{k + a_n} \) is also greater than 0, keeping the sequence above zero.
• We also showed \( a_1 = \sqrt{k} < \sqrt{2k} \). By repeating the process, we ensured every \( a_n < \sqrt{2k} \).

It's like the sequence is in a safe zone, with 0 and \( \sqrt{2k} \) as its boundaries. A bounded sequence helps in establishing other properties, like convergence.

Increasing Sequences

An increasing sequence is one where each term is greater than or equal to the one before it. For our sequence \( \{a_n\} \), we established it's strictly increasing.

We used induction to prove this:

• Initially, we see that \( a_1 = \sqrt{k} \) and \( a_2 = \sqrt{k + \sqrt{k}} > \sqrt{k} = a_1 \).
• Assume this holds for some \( n \), meaning \( a_n < a_{n+1} \).
• We then show \( a_{n+1} < a_{n+2} \) since \( a_{n+2} = \sqrt{k + a_{n+1}} > \sqrt{k + a_n} = a_{n+1} \).

Through this pattern, the sequence continually increases. Knowing a sequence is increasing allows us to make predictions about its behavior as it progresses.

Monotone Convergence Theorem

The Monotone Convergence Theorem is a crucial principle in calculus. It states that every bounded, monotonic (either entirely non-increasing or non-decreasing) sequence is convergent. This means it approaches a limit as \( n \) goes to infinity.

Applying this theorem:

• We've established that \( \{a_n\} \) is both bounded and increasing, thus fulfilling the conditions of the theorem.
• Therefore, by the theorem, \( \lim_{n \to \infty} a_n \) exists.

This theorem is a powerful tool. It gives certainty that sequences like ours, meeting specific conditions, will indeed converge to a limit. For our sequence, the limit represents the value that the terms are getting closer to as \( n \) becomes very large.