Question

If \(s_{1}=\sqrt{2}\), and $$s_{n+1}=\sqrt{2+\sqrt{s_{n}}} \quad(n=1,2,3, \ldots)$$ prove that \(\left\\{s_{n}\right\\}\) converges, and that \(s_{n}<2\) for \(n=1,2,3, \ldots .\)

Short answer

Question: Prove that the sequence \(s_n\) converges and that \(s_n<2\) for all \(n=1,2,3,\ldots\), where \(s_1=\sqrt{2}\) and \(s_{n+1}=\sqrt{2+\sqrt{s_n}}\). Answer: The sequence \(s_n\) converges since it is both monotonic increasing and bounded above. Additionally, through mathematical induction, we have shown that \(s_n<2\) for all \(n=1,2,3,\ldots\).

Step-by-step solution

Prove the sequence is monotonic

Let's show that the sequence is increasing. For simplicity, let \(x = s_n\). Then \(s_{n+1} = \sqrt{2+\sqrt{x}}\). We want to show that, for any \(n\), \(s_{n} \leq s_{n+1}\). To do that, let's compare \(s_{n}\) and \(s_{n+1}\): $$s_n = \sqrt{2+\sqrt{s_{n-1}}} \leq \sqrt{2+\sqrt{s_{n}}} = s_{n+1}$$ To prove that, we need to demonstrate that the function \(g(x) = \sqrt{2+\sqrt{x}}\) is increasing for the values of \(x\) in our sequence. Let's take the derivative of the function \(g(x)\): $$\frac{d}{dx}g(x) = \frac{d}{dx}\sqrt{2+\sqrt{x}} = \frac{1}{4\sqrt{(2+\sqrt{x})^3}}$$ We can see that the derivative is always positive for \(x\geq 0\). Thus, the function \(g(x)\) is increasing for any given \(x\) in our sequence. Therefore, the sequence is monotonic (increasing).

Prove the sequence is bounded

Next, let's show that the sequence is bounded. We want to prove that there exists an upper bound \(M\) such that \(s_n \leq M\) for all \(n \geq 1\). We have: $$s_1 = \sqrt{2} < 2$$ Assume that \(s_n < 2\). Then, $$s_{n+1} = \sqrt{2+\sqrt{s_n}} < \sqrt{2+\sqrt{2}}$$ Now, let's show that \(\sqrt{2+\sqrt{2}} < 2\). Squaring both sides of the inequality, we get: $$2+\sqrt{2} < 4$$ $$\sqrt{2} < 2$$ which is true. Therefore, \(s_{n+1} < 2\). Hence, the sequence is bounded above by 2.

Use the Monotone Convergence Theorem

Since we have shown that the sequence \(s_n\) is both monotonic increasing and bounded above, we can apply the Monotone Convergence Theorem. This theorem states that a sequence that is monotonic and bounded converges to a limit. Thus, the sequence \(s_n\) converges.

Prove \(s_n

We already showed that the sequence \(s_n\) is bounded above by 2. Now, we need to prove that \(s_n<2\) for all \(n=1,2,3,\ldots\). We can use mathematical induction to prove this statement. Base case: For \(n=1\), \(s_1=\sqrt{2}<2\). So, the base case holds. Inductive step: Assume that \(s_n<2\) for some \(n\geq 1\). Then, as we previously demonstrated, $$s_{n+1} = \sqrt{2+\sqrt{s_n}} < \sqrt{2+\sqrt{2}} < 2.$$ Thus, by mathematical induction, \(s_n<2\) for all \(n=1,2,3,\ldots\). In conclusion, the sequence \(s_n\) converges, and \(s_n<2\) for all \(n=1,2,3,\ldots\).

Key concepts

Mathematical Sequences

A mathematical sequence is essentially an ordered list of numbers that often follows a specific rule or pattern. For instance, consider the sequence where each term is defined recursively by:

\[s_{n+1} = \sqrt{2+\sqrt{s_n}}\]
In this example, the initial term, \(s_1\), is \(\sqrt{2}\), and subsequent terms are formed by the rule given. Sequences can be finite or infinite, and depending on their characteristics, they can be categorized in different ways. Recognizing the nature of the sequence is pivotal for understanding its behavior, such as whether it converges or diverges.

One relevant kind of sequence that students often encounter is a monotonic sequence, which either never increases or never decreases. The sequence described above is shown to be monotonic by proving that each term is less than or equal to the next term. Since it always increases (or stays the same), it is specifically called a monotonically increasing sequence. This distinction is essential for predicting the behavior of the sequence as it progresses and is a step toward proving convergence using the Monotone Convergence Theorem.

Convergence of Sequences

The concept of convergence in mathematical sequences is a vital one, referring to the idea that as you move through the sequence, the terms become arbitrarily close to a certain number called the limit.

Specifically, a sequence \(\left\{s_n\right\}\) is said to converge to a limit \(L\) if for every positive number \(\epsilon\), no matter how small, there exists some point in the sequence beyond which all terms of the sequence are within \(\epsilon\) of \(L\). Symbolically, we represent this by:

\[\lim_{n \rightarrow \infty}s_n = L\]
If you can find such a limit \(L\), then you can say the sequence converges to \(L\). To demonstrate convergence in the given sequence, we utilized the Monotone Convergence Theorem. As the name suggests, this theorem implies that if a sequence is monotonic and bounded, it must converge. After proving the sequence was both increasing and bounded above by 2, we utilized this theorem to conclude that the sequence indeed converges to some limit.

Bounded Sequences

A sequence is considered to be bounded if its terms stay within a fixed distance from a particular value, not venturing off towards infinity. More formally, a sequence \(\left\{s_n\right\}\) is said to be bounded above if there is some number \(M\) such that:

\[s_n \leq M\]
for all \(n\) in the sequence. If such an \(M\) exists, we can reassure ourselves that no matter how far out we go in the sequence, its values will never exceed \(M\).

Similarly, a sequence is bounded below if there's a number \(m\) serving as a lower limit. In the given problem, we proved that the sequence \(\left\{s_n\right\}\) was bounded above by 2, fulfilling the criteria necessary to use the Monotone Convergence Theorem. By demonstrating that \(s_n<2\) for all \(n\), we essentially showed the sequence will forever remain under this upper boundary, thus making it a bounded sequence. The bounded nature of sequences is a fundamental concept that intertwines with their convergence, as seen in the resolution of the exercise.