5. Show that the sequence {xn} defined by x1 =
√
7 and xn+1 =
√
7 + xn converges to the positive root of
the equation x
2 − x − 7 =

Let's solve the problem by showing that the sequence \(\{x_n\}\) defined by \(x_1 = \sqrt{7}\) and \(x_{n+1} = \sqrt{7 + x_n}\) converges to the positive root of the equation \(x^2 - x - 7 = 0\).

### Step 1: Find the roots of the equation
The given equation is:
\[
x^2 - x - 7 = 0
\]

To find the roots, use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = -1\), and \(c = -7\). Plugging in these values:
\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-7)}}{2(1)}
\]
\[
x = \frac{1 \pm \sqrt{1 + 28}}{2}
\]
\[
x = \frac{1 \pm \sqrt{29}}{2}
\]

So the roots are:
\[
x = \frac{1 + \sqrt{29}}{2} \quad \text{and} \quad x = \frac{1 - \sqrt{29}}{2}
\]

Since \(\sqrt{29} > 1\), the positive root is:
\[
\alpha = \frac{1 + \sqrt{29}}{2}
\]

### Step 2: Show that the sequence converges to \(\alpha\)
We need to show that \(\lim_{n \to \infty} x_n = \alpha\).

Assume that the sequence \(\{x_n\}\) converges to some limit \(L\). Then, by the definition of the sequence:
\[
L = \sqrt{7 + L}
\]

Square both sides:
\[
L^2 = 7 + L
\]

Rearranging terms gives:
\[
L^2 - L - 7 = 0
\]

This is exactly the quadratic equation we solved earlier, so \(L\) must be one of the roots of the equation. Since \(x_1 = \sqrt{7}\) is positive, and each \(x_n\) is defined as a square root (which is positive), the sequence \(\{x_n\}\) is positive, so the limit \(L\) must be the positive root:
\[
L = \frac{1 + \sqrt{29}}{2}
\]

### Step 3: Check if the sequence is increasing and bounded
Let's check whether the sequence \(\{x_n\}\) is increasing and bounded above by \(\alpha = \frac{1 + \sqrt{29}}{2}\).

#### Show that the sequence is increasing:
We will show that \(x_{n+1} > x_n\) for all \(n\).

Assume \(x_n < \alpha\). Then:
\[
x_{n+1} = \sqrt{7 + x_n}
\]
Since \(f(x) = \sqrt{7 + x}\) is an increasing function (because its derivative \(f'(x) = \frac{1}{2\sqrt{7+x}} > 0\)), and \(x_n < \alpha\), it follows that:
\[
x_{n+1} = \sqrt{7 + x_n} > \sqrt{7 + \alpha} = \alpha
\]
So the sequence is increasing.

#### Show that the sequence is bounded:
Next, we check if the sequence is bounded above by \(\alpha\).

If \(x_n \leq \alpha\), then:
\[
x_{n+1} = \sqrt{7 + x_n} \leq \sqrt{7 + \alpha} = \alpha
\]

Thus, the sequence is bounded above by \(\alpha\).

### Conclusion
Since the sequence \(\{x_n\}\) is increasing and bounded above, by the Monotone Convergence Theorem, it converges. The limit of the sequence must satisfy the equation \(L^2 - L - 7 = 0\), and since the sequence is positive, the limit is the positive root \(\alpha = \frac{1 + \sqrt{29}}{2}\).

Thus, the sequence \(\{x_n\}\) converges to the positive root of the equation \(x^2 - x - 7 = 0\).

---

Would you like any further details or explanations?

Here are some related questions you might find helpful:
1. What is the Monotone Convergence Theorem?
2. How do you determine if a sequence is bounded?
3. What are the general properties of quadratic equations?
4. Can a sequence have more than one limit?
5. How does the concept of limits apply in real analysis?
6. What are other methods to solve quadratic equations besides the quadratic formula?
7. How can you determine the rate of convergence of a sequence?


Explore the convergence of the sequence {xn} where x1 = √7 and xn+1 = √(7 + xn). Learn how this sequence converges to the positive root of the equation x^2 - x - 7 = 0 through detailed steps and the application of the Monotone Convergence Theorem. Suitable for advanced high school and undergraduate students studying sequences and limits.

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