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Suppose we have $x_1=\sqrt{2}$ and we have $x_{n+1}=(\sqrt{2})^{x_n}$. We have to find out the limit of the sequence.

I was observing that $x_2=(\sqrt{2})^\sqrt{2}$ and $x_3=(\sqrt{2})^{{\sqrt{2}}^{\sqrt{2}}}$ and so on now it is clear that the function is increasing. Now how to prove that the sequence is convergent and then how to find the limit? The calculative value is going towards $2$. But how to prove it!!

Ri-Li
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    You have $x_n < 2$ for all n. – GAVD May 17 '19 at 11:33
  • If the limit $x$ exists, then $x = \sqrt{2}^x$. Can you guess what $x$ is? – Paul Frost May 17 '19 at 11:35
  • After you've proven its existence, the limit fulfills the relation $y=\sqrt{2} ^y$ – B.Swan May 17 '19 at 11:36
  • Looks like tetration. – Wuestenfux May 17 '19 at 12:01
  • Define the function $f:\Bbb [0,2] \to [0,2]: x\mapsto\sqrt 2^x$. $f$ is smooth. Note that $f'(x)=\sqrt2^x\cdot\ln(\sqrt2)<1$ for all $x\in[0,2]$. Thus $f$ is a contraction. Note that $2$ is the only fixed point of $f$.

    By the Banach fixed point Theorem, any sequence of the type $x_0\in[0,2], x_{n+1}:=f(x_n)$ converges to the fixed point of $f$. In this case $2$.

     – Maximilian Janisch Dec 22 '19 at 20:06

3 Answers3

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  1. To prove that the limit exists, you can prove, by induction, that $x_n<2$ is true for all $n$. From this, and the fact that the sequence is increasing, it follows that it must have a limits.
  2. To find the limit, think about what happens to the equality $x_{n+1}=\sqrt{2}^{x_n}$ when $n\to\infty$. Remember that $$\lim_{n\to\infty} x_n=\lim_{n\to\infty}x_{n+1}$$
5xum
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The limit of that sequence, if it exists, will be a fixed point of $g(x) = \sqrt{2}^x$. So, if you solve the equation $x = \sqrt{2}^x$, you have all possible limits (the solutions are 2 and 4). Ultimately, you should use the fixed point theorem to establish convergence when you start with $x_1 = \sqrt{2}$.

Just as a curiosity, the other fixed point, $x=4$, can only be reached if you set $x_1=4$. Otherwise, if $x_1 < 4$ the sequence will converge to $2$ and if $x_1>4$ the sequence will diverge.

PierreCarre
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If $x<2$, then $\sqrt 2^x<2$ (the exponential is a growing function). You have a bounded increasing sequence, so it converges.

Upon convergence, $x=\sqrt2^x$, or $2\ln_2 x-x=0$. The derivative of the LHS is $\dfrac{2}{x\ln2}-1$, which has a single root, hence the function has at most two roots. By inspection, they are $2$ and $4$.

The iterations from $x_1=1$ do converge to $x=2$.