# Prove that the sequence $\sqrt{2}, {\sqrt{2}}^{\sqrt{2}}$, … converges, showing that it is increasing and bounded above. Find your limit.

$$a_1=\sqrt{2}$$ $$a_2={\sqrt{2}}^{\sqrt{2}}={\sqrt{2}}^{a_1}$$ $$a_3={\sqrt{2}}^{{\sqrt{2}}^{\sqrt{2}}}={\sqrt{2}}^{a_2}$$ $$.$$ $$.$$ $$.$$ $$a_{n+1}={\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}}^{{.}^{{.}^{.}}}}}={\sqrt{2}}^{a_n}$$ We will see that the secuence is increasing. Note that. $$a_1=\sqrt{2}<{\sqrt{2}}^{\sqrt{2}}=a_2$$ Suppose that $$a_{k+1}={\sqrt{2}}^{a_k}<{\sqrt{2}}^{a_{k+1}}=a_{k+2}$$ Applying $log_{\sqrt{2}}$ $$log_{\sqrt{2}}({{\sqrt{2}}^{a_k}})<log_{\sqrt{2}}({{\sqrt{2}}^{a_{k+1}}})$$ $$\Rightarrow$$ $$a_k{log_{\sqrt{2}}({\sqrt{2}}})<a_{k+1}{log_{\sqrt{2}}({\sqrt{2}}})$$ Thus $$a_k<a_{k+1}$$ We will see that the sequence is bounded above. Note that $$a_1=\sqrt{2}<2$$ $$a_2={\sqrt{2}}^{a_1}<{\sqrt{2}}^{2}=2$$ Suppose that
$$a_k<2$$ then $$a_{k+1}={\sqrt{2}}^{a_k}<{\sqrt{2}}^{2}=2$$ Therefore, the sequence is bounded above by 2.

We will see that $lim a_n=2$. Be $$lim a_n=L \Rightarrow lim a_{n+1}=L$$ We have that $a_{n+1}={\sqrt{2}}^{a_n}$

$\Rightarrow$ $L={\sqrt{2}}^{L}$

Applying $log_2$ $$log_{2}(L)=log_2(2^{l/2})$$ $$\Rightarrow 2=L/log_2(L)$$ We have two cases

i) L=4 since $4/log_2(4)=2$

ii) L=2 since $2/log_2(2)=2$

Since the sequence is bounded above by 2 so we concluded that L=2. Do you think the proof is correct? Any other way to prove it?

Hint

Let $f(x)=2^\frac x2=e^{\frac x2 \ln(2)}$

we have that

$$(\forall n>0) \;\;\;a_{n+1}=f(a_n).$$

$f$ is increasing and $a_2>a_1$

$\implies (a_n)_n$ is increasing.

the fixe point of $f$ is such that

$f(L)=L=2^\frac L2$

$\implies L=2\;$or$\;L=4$

By induction, we prove easily that $$\forall n\geq 1\;\; 0<a_n\leq 2$$

finally $(a_n)_n$ as increasing and bounded, converges to $2$.

If there is convergence to $a$, we must have $$a-\sqrt2^a=0.$$By inspection, $a=2$ and $a=4$ are solutions. Taking the derivative, the LHS has a single root $b$ such that $$1-\sqrt2^b\ln\sqrt2=0.$$ It verifies $2<b<4$ and there are no others solution.

Then, for $x<2$, we have $$x<\sqrt2^x<2,$$ so that the iterates are strictly increasing and bounded above.

As the starting iterate is $\sqrt2$, there is no need to care about the other solution.