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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.

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  • $\begingroup$ You can use the monotone convergence theorem $\endgroup$ – Dr. Sonnhard Graubner Dec 10 '18 at 7:48
  • $\begingroup$ @Dr. Sonnhard Graubner but how to prove 2 is least upper bound? $\endgroup$ – ramanujan Dec 10 '18 at 7:53
  • $\begingroup$ Taking $\log$, then taking the limit $n \to \infty$. $\endgroup$ – xbh Dec 10 '18 at 7:53
  • $\begingroup$ You can use induction to show that $$x_n<2$$ $\endgroup$ – Dr. Sonnhard Graubner Dec 10 '18 at 7:54
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    $\begingroup$ $l=4$ is also a solution $\endgroup$ – Masacroso Dec 10 '18 at 8:22
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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 $[1,2]\Leftrightarrow x=2$
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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.

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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.

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