# If $x_1=\sqrt 2$ and $x_{n+1}=(\sqrt2)^{x_n}$ then sequence $x_n$ converges to 2.

If $$x_1=\sqrt 2$$ and $$x_{n+1}=(\sqrt2)^{x_n}$$ then show that sequence $$x_n$$ converges to 2.

I know this sequence is monotonically increasing. But how to prove it converges to 2?

The sequence is bounded above is already answered here. And I know that$$\sqrt 2 ^l=l$$ has only solution 2. But how to prove formally? Thanks.

• You can use the monotone convergence theorem – Dr. Sonnhard Graubner Dec 10 '18 at 7:48
• @Dr. Sonnhard Graubner but how to prove 2 is least upper bound? – ramanujan Dec 10 '18 at 7:53
• Taking $\log$, then taking the limit $n \to \infty$. – xbh Dec 10 '18 at 7:53
• You can use induction to show that $$x_n<2$$ – Dr. Sonnhard Graubner Dec 10 '18 at 7:54
• $l=4$ is also a solution – Masacroso Dec 10 '18 at 8:22

## 3 Answers

Here is a formal way considering $$f(x) = \left( \sqrt{2}\right)^x$$ and using MVP:

• On $$[1,2]$$ we have $$0 < f'(x)= \ln{\sqrt{2}}\left( \sqrt{2}\right)^x \leq 2 \ln{\sqrt{2}} < 2 \cdot \frac{2}{5}= \frac{4}{5}$$ $$0 \leq 2 - x_{n+1} = \left( \sqrt{2}\right)^2 - \left( \sqrt{2}\right)^{x_n} = f'(\xi_n)(2-x_n) < \frac{4}{5}(2-x_n)$$

It follows $$0 \leq 2 - x_{n+1} < \left(\frac{4}{5}\right)^n (2-x_1)\stackrel{n \to \infty}{\longrightarrow} 0 \Rightarrow \boxed{\lim_{n \to \infty}x_n = 2}$$

To sum this up:

• $$f$$ maps $$[1,2]$$ into $$[1,2]$$ and, hence, has a fixpoint there.
• The fixpoint is unique as $$|f'(x)| \leq q < 1$$ on $$[1,2]$$.
• For any starting value $$x_1 \in [1,2]$$ the iteration $$x_{n+1} = f(x_n)$$ will converge to the fixpoint which is the solution of $$x = f(x)$$ on $$[1,2]\Leftrightarrow x=2$$

To find a candidate for the limit you have the equation

$$x=(\sqrt 2)^x\tag1$$

for some possible limit $$x:=\lim_{n\to\infty}x_n$$. This gives, assuming $$x>0$$, the equivalent equation $$x^{1/x}=\sqrt 2$$, what have the possible solutions $$2$$ and $$4$$, what can be seen by the study of the function $$x^{1/x}$$.

Thus, if you shows that the sequence is monotone and bounded above by $$2$$ you are done.

REMARK: you dont need to show that $$2$$ is the least upper bound, just that it is an upper bound, because the unique possible solutions are $$2$$ and $$4$$, and by the monotony of the sequence we knows that the sequence converges if it is bounded.

As the exponential is a growing function,

$$x<2\implies \sqrt2^{\,x}<\sqrt2^{\,2}=2$$ and the sequence starting from $$x=\sqrt2<2$$ is bounded above.