Solutions

(a) By rewriting the sequence as a recursive one,

$$y_{n} = (\sqrt{2}, \; \sqrt{2 + \sqrt{2}},\; \sqrt{2 + \sqrt{2 + \sqrt{2}}}, \; \dots)$$

is equivalent to

$$y_{n + 1} = \sqrt{2 + y_{n}}$$

with $y_1 = \sqrt{2}$. This makes it easier for us to prove the conditions of MCT, which we attempt, to demonstrate that the sequence converges.

Since the square root function is increasing, we know that the sequence also is increasing. Next, we prove that the sequence is bounded above. Let us try to use the upper bound 2, solely because 2 is one of the recurring arguments of the sequence. The base case is obviously true since

$$y_{1} = \sqrt{2} < 2.$$

For the induction step, assume that $y_{n} < 2$. Since the sequence is increasing and the lower bound is $y_1 = \sqrt{2}$, all terms in the sequence are positive. Then,

$$\begin{aligned} y_{n + 1} = \sqrt{2 + y_{n}} < 2 &\iff 2 + y_{n} < 4 \\ &\iff y_{n} < 2. \end{aligned}$$

Now knowing that the limit exists, we apply the strategy shown in the previous exercise to find it. By labeling $\lim_{n \to \infty} y_{n}$ as $L$, we have

$$\begin{aligned} \lim_{n \to \infty} y_{n} = \lim_{n \to \infty} \sqrt{2 + y_{n}} &\implies L = \sqrt{2 + L} \\ &\implies L^2 - L - 2 = 0 \\ &\implies (L + 1)(L - 2) = 0 \\ &\implies x_1 = -1, \quad x_2 = 2. \end{aligned}$$

Since the sequence is always positive, the limit is $x_2 = 2$.

(b) As for (a), we rewrite the sequence as a recursive one, rendering

$$y_{n} = (\sqrt{2}, \; \sqrt{2 \sqrt{2}}, \; \sqrt{2 \sqrt{2 \sqrt{2}}}, \; \dots)$$

as

$$y_{n + 1} = \sqrt{2 y_{n}}$$

with $y_1 = 2$. Again, we first prove that the sequence converges before applying the now common strategy to find the limit of the sequence. Since this exercise has a close resemblance to (a), I leave it as an exercise. The limit is 2, as in (a).