Contractive Sequence: $x_{n+1}=2+\frac{1}{x_n}$ Convergence

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Hey everyone! Today, we're diving deep into the fascinating world of iterative sequences, specifically the sequence defined by $x_{n+1}=2+\frac{1}{x_n}$, with the initial term $x_1 = 2$. Our main goal is to show that this sequence is contractive, which automatically implies that it's convergent. And, as a bonus, we'll also figure out what its limit is. So, buckle up, and let's get started!

Understanding Contractive Sequences

Before we jump into the specifics of our sequence, let's quickly recap what it means for a sequence to be contractive. In simple terms, a sequence is contractive if the terms get closer and closer to each other as we move along the sequence. More formally, a sequence $(x_n)$ is contractive if there exists a constant $0 \le k < 1$ such that

$|x_{n+2} - x_{n+1}| \le k |x_{n+1} - x_n|$

for all $n \ge 1$. The constant $k$ is often referred to as the contraction constant. This inequality essentially states that the distance between consecutive terms decreases by a factor of at least $k$ at each step. This "shrinking" property is what guarantees that the sequence converges to a limit.

Why is being contractive so important? Well, contractive sequences are guaranteed to converge to a limit within the real numbers. This is a powerful result, as it gives us a way to prove convergence without explicitly knowing the limit beforehand. The Banach Fixed-Point Theorem formalizes this idea, stating that every contractive mapping in a complete metric space (like the real numbers) has a unique fixed point, which is the limit our sequence will approach.

When dealing with iterative sequences like ours, proving contractiveness often involves some algebraic manipulation and careful estimation. We need to show that the distance between successive terms shrinks sufficiently, and this usually requires us to bound some expression involving the terms of the sequence.

In our specific case, the function that defines our sequence is $f(x) = 2 + \frac{1}{x}$. To show that the sequence is contractive, we'll need to look at the difference between $f(x_{n+1})$ and $f(x_n)$ and relate it to the difference between $x_{n+1}$ and $x_n$. This will involve using the properties of the function $f$ and the nature of our initial term.

Now that we have a good grasp of what contractiveness means, let's dive into the nitty-gritty details of our sequence!

Proving the Sequence is Contractive

Okay, let's get our hands dirty and show that the sequence $x_{n+1}=2+\frac{1}{x_n}$ is indeed contractive. Remember, we need to find a constant $0 \le k < 1$ such that $|x_{n+2} - x_{n+1}| \le k |x_{n+1} - x_n|$ for all $n \ge 1$.

First, let's express $x_{n+2}$ and $x_{n+1}$ in terms of the previous terms:

$x_{n+1} = 2 + \frac{1}{x_n}$ and $x_{n+2} = 2 + \frac{1}{x_{n+1}}$

Now, let's look at the difference between $x_{n+2}$ and $x_{n+1}$:

$|x_{n+2} - x_{n+1}| = |(2 + \frac{1}{x_{n+1}}) - (2 + \frac{1}{x_n})| = |\frac{1}{x_{n+1}} - \frac{1}{x_n}|$

To simplify this further, let's find a common denominator:

$|\frac{1}{x_{n+1}} - \frac{1}{x_n}| = |\frac{x_n - x_{n+1}}{x_n x_{n+1}}|$

Now, we can rewrite this as:

$|x_{n+2} - x_{n+1}| = \frac{|x_{n+1} - x_n|}{|x_n x_{n+1}|}$

This is a crucial step! We've managed to express the difference between $x_{n+2}$ and $x_{n+1}$ in terms of the difference between $x_{n+1}$ and $x_n$, which is exactly what we wanted. Now, we need to find a bound for the term $\frac{1}{|x_n x_{n+1}|}$.

To do this, we need to understand the behavior of the sequence $(x_n)$. Let's calculate the first few terms:

• $x_1 = 2$
• $x_2 = 2 + \frac{1}{2} = 2.5$
• $x_3 = 2 + \frac{1}{2.5} = 2.4$
• $x_4 = 2 + \frac{1}{2.4} \approx 2.4167$

It seems like the sequence is oscillating and converging. A crucial observation here is that the terms appear to be bounded below by 2. We can prove this rigorously using induction.

Base Case: For $n = 1$, $x_1 = 2 \ge 2$.

Inductive Step: Assume $x_n \ge 2$ for some $n \ge 1$. Then,

$x_{n+1} = 2 + \frac{1}{x_n} \ge 2 + \frac{1}{\infty}$ (As $x_n$ increase, $1/x_n$ decrease and approach 0) = 2+0 = 2

So, by induction, $x_n \ge 2$ for all $n \ge 1$. This is fantastic news! It gives us a lower bound for the terms of our sequence.

Now, we can use this bound to estimate the term $\frac{1}{|x_n x_{n+1}|}$. Since both $x_n$ and $x_{n+1}$ are greater than or equal to 2, we have:

$|x_n x_{n+1}| = x_n x_{n+1} \ge 2 * 2 = 4$

Therefore,

$\frac{1}{|x_n x_{n+1}|} \le \frac{1}{4}$

Plugging this back into our inequality, we get:

$|x_{n+2} - x_{n+1}| = \frac{|x_{n+1} - x_n|}{|x_n x_{n+1}|} \le \frac{1}{4} |x_{n+1} - x_n|$

Eureka! We've found our contraction constant! We have shown that $|x_{n+2} - x_{n+1}| \le k |x_{n+1} - x_n|$ with $k = \frac{1}{4}$, which is indeed less than 1. This proves that the sequence is contractive. Awesome!

Finding the Limit of the Sequence

Now that we've successfully demonstrated that our sequence is contractive, we know it converges. But what does it converge to? Let's find the limit.

Let's assume that the limit of the sequence $(x_n)$ exists and is equal to $L$. That is, $\lim_{n \to \infty} x_n = L$. Since $x_{n+1}$ is just the next term in the sequence, it should also converge to the same limit. So, we have $\lim_{n \to \infty} x_{n+1} = L$ as well.

Now, let's take the limit of both sides of the recursive definition of our sequence:

$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} (2 + \frac{1}{x_n})$

Using the limit laws, we can rewrite this as:

$L = 2 + \frac{1}{L}$

This is a simple equation that we can solve for $L$. Multiplying both sides by $L$, we get:

$L^2 = 2L + 1$

Rearranging the terms, we get a quadratic equation:

$L^2 - 2L - 1 = 0$

We can solve this equation using the quadratic formula:

$L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$

So, we have two possible values for the limit: $1 + \sqrt{2}$ and $1 - \sqrt{2}$. However, we know that $x_n \ge 2$ for all $n$, so the limit must also be greater than or equal to 2. Since $1 - \sqrt{2}$ is negative, we can discard it. Therefore, the limit of the sequence is:

$L = 1 + \sqrt{2}$

Conclusion: Triumphant Convergence!

Woohoo! We've successfully shown that the iterative sequence $x_{n+1}=2+\frac{1}{x_n}$, with $x_1 = 2$, is contractive and converges to the limit $1 + \sqrt{2}$. We started by understanding the concept of contractive sequences and the Banach Fixed-Point Theorem. Then, we went through the steps of proving contractiveness by finding a suitable contraction constant. Finally, we calculated the limit by solving a simple equation.

This journey demonstrates the power of mathematical analysis in understanding the behavior of sequences and their convergence. The concepts we've explored here, like contractive mappings and fixed points, have far-reaching applications in various fields, including numerical analysis, optimization, and differential equations.

So, the next time you encounter an iterative sequence, remember the tools and techniques we've discussed today. You might just be able to unravel its secrets and find its limit!