To prove that the sequence $(x_n)$, defined by $x_1 = \sqrt{2}$ and $x_{n+1} = \sqrt{2x_n}$ for $n \geq 1$, converges and that $\lim_{n \to \infty} x_n = 2$, we will follow these steps:

1. Assume the limit exists

Let us assume that the sequence $(x_n)$ converges to a limit $L$. Then, as $n \to \infty$, both $x_n$ and $x_{n+1}$ converge to the same limit $L$. Therefore, taking the limit on both sides of the recurrence relation:

$L = \sqrt{2L}$

2. Solve for $L$

Squaring both sides of the equation:

$L^2 = 2L$

Rearranging this:

$L^2 - 2L = 0$

Factoring:

$L(L - 2) = 0$

This gives two possible solutions: $L = 0$ or $L = 2$.

3. Eliminate the case $L = 0$

Since $x_1 = \sqrt{2} > 0$ and all terms of the sequence are defined as square roots, which are positive, the sequence $(x_n)$ consists of strictly positive terms. Therefore, $L = 0$ is not possible. Hence, the only valid solution is:

$L = 2$

Thus, if the sequence converges, it must converge to 2.

4. Prove monotonicity and boundedness

To complete the proof, we need to show that the sequence is both bounded and monotonic, which will guarantee convergence by the Monotone Convergence Theorem.

a) Boundedness

We will show that the sequence $(x_n)$ is bounded above by 2. Assume $x_1 = \sqrt{2} < 2$. Now, by induction, assume $x_n < 2$. From the recurrence relation:

$x_{n+1} = \sqrt{2x_n}$

Since the square root function is increasing and $x_n < 2$, we have:

$x_{n+1} = \sqrt{2x_n} < \sqrt{2 \times 2} = 2$

Therefore, by induction, $x_n < 2$ for all $n$. Thus, $(x_n)$ is bounded above by 2.

b) Monotonicity

Next, we show that the sequence is increasing. We need to prove that $x_{n+1} \geq x_n$. That is, we need to verify:

$\sqrt{2x_n} \geq x_n$

Squaring both sides:

$2x_n \geq x_n^2$

Rearranging:

$x_n^2 - 2x_n \leq 0$

Factoring:

$x_n(x_n - 2) \leq 0$

Since $x_n > 0$ and $x_n < 2$, the product $x_n(x_n - 2)$ is non-positive, so the inequality holds. Therefore, $x_{n+1} \geq x_n$, meaning the sequence is increasing.

5. Conclusion

The sequence $(x_n)$ is bounded above by 2 and is increasing. By the Monotone Convergence Theorem, the sequence converges. Since we already found that the only possible limit is 2, we conclude that:

$\lim_{n \to \infty} x_n = 2$

Would you like more details on any part of the solution?

5 Related Questions

1. Can you prove the convergence of sequences using other methods like the Cauchy criterion?
2. What is the significance of the Monotone Convergence Theorem?
3. How would the behavior of the sequence change if the initial value $x_1$ were different?
4. Can you generalize the recurrence relation $x_{n+1} = \sqrt{2x_n}$ to other constants besides 2?
5. What are some examples of divergent sequences, and how do they differ from this convergent sequence?

Tip

When proving the convergence of a sequence, always consider both boundedness and monotonicity if applicable. These two properties are key in applying the Monotone Convergence Theorem.