Exercise 2.4.3

Question

\begin{enumerate}[label=(\alph*)] 
  \item Show that
    $$
    \sqrt{2}, \sqrt{2+\sqrt{2}}, \sqrt{2+\sqrt{2+\sqrt{2}}}, \ldots
    $$
    converges and find the limit.
  \item Does the sequence
    $$
    \sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, \ldots
    $$
    converge? If so, find the limit.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Let $x_1 = \sqrt 2$ and $x_{n+1} = \sqrt{2 + x_n}$ clearly $x_2 > x_1$. assuming $x_{n+1} > x_n$ gives
    $$
    2 + x_{n+1} > 2 + x_n \iff \sqrt{2 + x_{n+1}} > \sqrt{2 + x_n} \iff x_{n+2} > x_{n+1}
    $$
    Since $x_n$ is monotonically increasing and bounded the monotone convergence theorem tells us $(x_n) 	o x$. Equating both sides like in 2.4.1 gives
    $$
    x = \sqrt{2 + x} \iff x^2 - x - 2 = 0 \iff x = \frac 12 \pm \frac{3}{2}
    $$
    Since $x > 0$ we must have $x = 2$.
  \item We have $x_1 = 2^{1/2}$ and $x_{n+1} = (2x_n)^{1/2}$. We have
    $$
    x_{n+1} = (2x_n)^{1/2} \ge x_n \iff 2x_n \ge x_n^2 \iff 2 \ge x_n
    $$
    Since $x_1 = 2^{1/2} \le 2$ induction implies $x_n$ is increasing. Now to show $x_n$ is bounded notice that $x_1 \le 2$ and if $x_n \le 2$ then
    $$
    2x_n \le 4 \implies (2x_n)^{1/2} \le 2
    $$
    Now the monotone convergence theorem tells us $(x_n)$ converges. To find the limit use $\lim x_n = \lim x_{n+1} = x$ to get
    $$
    x = (2x)^{1/2} \implies x^2 = 2x \implies x = \pm 2
    $$
    Since $x_n \ge 0$ we have $x = 2$.
   \end{enumerate}

Exercise 2.4.3

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Exercise 2.4.3

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