Prove That The Sequence Defined By X N + 1 = X N + C X_{n+1}=\sqrt{x_n +c} X N + 1 ​ = X N ​ + C ​ Converges Monotonically

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Introduction

In this article, we will prove that the sequence defined by $x_{n+1}=\sqrt{x_n +c}$ converges monotonically. This sequence is defined recursively, where each term is the square root of the previous term plus a constant $c$. We will show that this sequence is monotonically increasing and bounded above, which implies that it converges.

The Sequence and Its Properties

The sequence $\{x_n\}$ is defined recursively as follows:

• $x_1 > 0$ is fixed
• For $n \geq 1$, $x_{n+1} = \sqrt{x_n + c}$

We are given that $c > 0$. We need to show that this sequence converges monotonically.

Monotonicity of the Sequence

To show that the sequence is monotonically increasing, we need to show that $x_{n+1} \geq x_n$ for all $n \geq 1$.

Let's assume that $x_n \geq x_{n-1}$ for some $n \geq 2$. We need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Substituting $x_n = \sqrt{x_{n-1} + c}$, we get:

$\sqrt{x_{n-1} + c} + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$\sqrt{x_{n-1} + c} \geq x_{n-1}$

Squaring both sides, we get:

$x_{n-1} + c \geq x_{n-1}^2$

Subtracting $x_{n-1}$ from both sides, we get:

$c \geq x_{n-1}^2 - x_{n-1}$

Factoring the right-hand side, we get:

$c \geq x_{n-1}(x_{n-1} - 1)$

Since $c > 0$, we have:

$x_{n-1}(x_{n-1} - 1) \leq 0$

This implies that $x_{n-1} \leq 1$.

Now, we need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Substituting $x_n = \sqrt{x_{n-1} + c}$, we get:

$\sqrt{x_{n-1} + c} + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$\sqrt{x_{n-1} + c} \geq x_{n-1}$

Squaring both sides, we get:

$x_{n-1} + c \geq x_{n-1}^2$

Subtracting $x_{n-1}$ from both sides, we get:

$c \geq x_{n-1}^2 - x_{n-1}$

Factoring the right-hand side, we get:

$c \geq x_{n-1}(x_{n-1} - 1)$

Since $c > 0$, we have:

$x_{n-1}(x_{n-1} - 1) \leq 0$

This implies that $x_{n-1} \leq 1$.

Since $x_{n-1} \leq 1$, we have:

$x_n = \sqrt{x_{n-1} + c} \geq \sqrt{1 + c}$

Since $x_n \geq \sqrt{1 + c}$, we have:

$x_{n+1} = \sqrt{x_n + c} \geq \sqrt{\sqrt{1 + c} + c}$

Since $\sqrt{1 + c} \geq 1$, we have:

$\sqrt{\sqrt{1 + c} + c} \geq \sqrt{1 + c}$

Since $x_{n+1} \geq \sqrt{1 + c}$, we have:

$x_{n+1} \geq x_n$

This shows that the sequence is monotonically increasing.

Boundedness of the Sequence

To show that the sequence is bounded above, we need to show that there exists a real number $M$ such that $x_n \leq M$ for all $n \geq 1$.

Let's assume that $x_n \geq x_{n-1}$ for some $n \geq 2$. We need to show that $x_n \leq M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + cSubtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_n \leq M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Substituting $x_n = \sqrt{x_{n-1} + c}$, we get:

$\sqrt{x_{n-1} + c} + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$\sqrt{x_{n-1} + c} \geq x_{n-1}$

Squaring both sides, we get:

$x_{n-1} + c \geq x_{n-1}^2$

Subtracting $x_{n-1}$ from both sides, we get:

$c \geq x_{n-1}^2 - x_{n-1}$

Factoring the right-hand side, we get:

$c \geq x_{n-1}(x_{n-1} - 1)$

Since $c > 0$, we have:

$x_{n-1}(x_{n-1} - 1) \leq 0$

This implies that $x_{n-1} \leq 1$.

Now, we need to show that $x_n \leq M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_n \leq M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

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Q: What is the sequence defined by $x_{n+1}=\sqrt{x_n +c}$?

A: The sequence $\{x_n\}$ is defined recursively as follows:

• $x_1 > 0$ is fixed
• For $n \geq 1$, $x_{n+1} = \sqrt{x_n + c}$

Q: What is the condition on $c$?

A: We are given that $c > 0$.

Q: How do we show that the sequence is monotonically increasing?

A: To show that the sequence is monotonically increasing, we need to show that $x_{n+1} \geq x_n$ for all $n \geq 1$.

Let's assume that $x_n \geq x_{n-1}$ for some $n \geq 2$. We need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Substituting $x_n = \sqrt{x_{n-1} + c}$, we get:

$\sqrt{x_{n-1} + c} + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$\sqrt{x_{n-1} + c} \geq x_{n-1}$

Squaring both sides, we get:

$x_{n-1} + c \geq x_{n-1}^2$

Subtracting $x_{n-1}$ from both sides, we get:

$c \geq x_{n-1}^2 - x_{n-1}$

Factoring the right-hand side, we get:

$c \geq x_{n-1}(x_{n-1} - 1)$

Since $c > 0$, we have:

$x_{n-1}(x_{n-1} - 1) \leq 0$

This implies that $x_{n-1} \leq 1$.

Now, we need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have$x_n + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_{n+1} \geq x_n$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Substituting $x_n = \sqrt{x_{n-1} + c}$, we get:

$\sqrt{x_{n-1} + c} + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$\sqrt{x_{n-1} + c} \geq x_{n-1}$

Squaring both sides, we get:

$x_{n-1} + c \geq x_{n-1}^2$

Subtracting $x_{n-1}$ from both sides, we get:

$c \geq x_{n-1}^2 - x_{n-1}$

Factoring the right-hand side, we get:

$c \geq x_{n-1}(x_{n-1} - 1)$

Since $c > 0$, we have:

$x_{n-1}(x_{n-1} - 1) \leq 0$

This implies that $x_{n-1} \leq 1$.

Since $x_{n-1} \leq 1$, we have:

$x_n = \sqrt{x_{n-1} + c} \geq \sqrt{1 + c}$

Since $x_n \geq \sqrt{1 + c}$, we have:

$x_{n+1} = \sqrt{x_n + c} \geq \sqrt{\sqrt{1 + c} + c}$

Since $\sqrt{1 + c} \geq 1$, we have:

$\sqrt{\sqrt{1 + c} + c} \geq \sqrt{1 + c}$

Since $x_{n+1} \geq \sqrt{1 + c}$, we have:

$x_{n+1} \geq x_n$

This shows that the sequence is monotonically increasing.

Q: How do we show that the sequence is bounded above?

A: To show that the sequence is bounded above, we need to show that there exists a real number $M$ such that $x_n \leq M$ for all $n \geq 1$.

Let's assume that $x_n \geq x_{n-1}$ for some $n \geq 2$. We need to show that $x_n \leq M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_n \le M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Substituting $x_n = \sqrt{x_{n-1} + c}$, we get:

$\sqrt{x_{n-1} + c} + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$\sqrt{x_{n-1} + c} \geq x_{n-1}$

Squaring both sides, we get:

$x_{n-1} + c \geq x_{n-1}^2$

Subtracting $x_{n-1}$ from both sides, we get:

$c \geq x_{n-1}^2 - x_{n-1}$

Factoring the right-hand side, we get:

$c \geq x_{n-1}(x_{n-1} - 1)$

Since $c > 0$, we have:

$x_{n-1}(x_{n-1} - 1) \leq 0$

This implies that $x_{n-1} \leq 1$.

Now, we need to show that $x_n \leq M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Subtracting $c$ from both sides, we get:

$x_n \geq x_{n-1}$

Now, we need to show that $x_n \leq M$ for some real number $M$.

Using the recursive definition of the sequence, we have:

$x_{n+1} = \sqrt{x_n + c}$

$x_n = \sqrt{x_{n-1} + c}$

Since $x_n \geq x_{n-1}$, we have:

$x_n + c \geq x_{n-1} + c$

Substituting $x_n = \