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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ $l=4$ is also a solution $\endgroup$ – Masacroso Dec 10 '18 at 8:22

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.