Show that a sequence involving $\sqrt2$ converges

Question

Show that the sequence

$$\sqrt2, \sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}} ,\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}},...$$

converges and find the limit

Answer 1

The sequence is given by $a_{n+1}=\sqrt{2a_n}$. Here is a roadmap:

• Prove that $a_n < 2$.

• Prove that $a_{n+1} > a_n$.

• Conclude that $(a_n)$ converges.

• If $L=\lim a_n$, then $L=\sqrt{2L}$.

Answer 2

Since $$x_n=2^{1-2^{-n}}\quad n=1,2,\ldots$$ Thus $\lim_{n\to\infty}x_n=2$.

Answer 3

Hint: Show that the sequence is increasing and use the fact that $\sqrt2\leq2$ to show $x_n\leq2$ for all $n$. This shows convergence, say to $x$. Then you have $\sqrt{2x}=x$ (why?) and you can solve this for $x$.

Answer 4

Define a sequence $(u_n)_{n \ge 0}$ by $u_0 = 1$ and $u_{n+1} = \sqrt{2 u_n}$.

We can show recursively that $u_n \le u_{n+1} \le 2$. So $(u_n)_{n \ge 0}$ converges. Let $l$ be the limit, we have that $l = \sqrt{2l}$, so $l = 2$.

Alternatively, define $v_n = \log_2(u_n)$. We have $v_0 = 0$ and $v_{n+1} = \frac{1+v_n}{2}$, so we can show $v_n = \sum_{k = 1}^n \frac{1}{2^k} \underset{n \to \infty}{\longrightarrow} 1$.