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Prove that following sequence converges and find its limit $$a_0 = \sqrt{2}$$ $$a_{n+1} = \sqrt{2 + \sqrt{a_n}}$$ My work: I know I have to prove that sequence is bounded and monotonous in order to prove that it converges, but I don't know how to prove both of it. I tried to prove that $a_n$ is bounded above by 2, and that is true, I got that using induction, but I don't know how to prove that $a_{n+1} - a_n > 0$ or $\frac{a_{n+1}}{a_n}>1$. Limit itself is not as big problem as to prove convergence, what I did is that, recurent equality becomes following, where L is limit $$L = \sqrt{2+ \sqrt{L}}$$ and from that we can easy get $$ L^2 - 2 = \sqrt{L}$$ where $L$ must be greater than $\sqrt{2}$. That is same as this equation $$(L - 1)(L^3 + L^2 -3L - 4)=0$$ and because $L$ can not be 1 then, using Cardano's formula I have got that $L$ is approximately $1.83$.

smth
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  • That is what I have mentioned above, problem is that I don't know how to prove that it is monotonous – smth Apr 19 '22 at 17:12
  • @OliverDíaz I know that problem, but it is a lot easier because, here, I have one root more and its harder to manipulate with – smth Apr 19 '22 at 17:13
  • Not orthodox - but one can also notice that if we look at this as a continuous function: $$f'(x) = \frac{1}{2\sqrt{2 + \sqrt x}} \frac{1}{2\sqrt x} > 0$$ – Gregory Apr 19 '22 at 17:18
  • @Gregory I'm not sure whether $f(x) = \sqrt{2+\sqrt{x}}$ being increasing is useful. More useful would be to prove $f(x) > x$ for $0 \le x < L$ where $L$ is the fixed point -- and for that, you can use the intermediate value theorem on $f(x) - x$ and a proof by contradiction. – Daniel Schepler Apr 19 '22 at 18:26
  • @DanielSchepler good point! – Gregory Apr 20 '22 at 13:18

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  • You can use numerical methods to find an approximate solution. The point is that the sequence $a_n$ is indeed convergent. Proving that it is bounded and that it has monotonicity properties is not complicated in this case. For example $a_1>a_0$ then, by induction, if $a_n>a_{n-1}$, then $$a_{n+1}=\sqrt{2+\sqrt{a_n}}>\sqrt{2+\sqrt{a_{n-1}}}=a_n$$ Boundedness can also be proven by induction.

  • The limit $L$, as you pointed out satisfies the quartic equation in your posting. Notice that if $u=\sqrt{L}$, then $u$ satisfies the quartic equation $$p(u)=u^4−u−2=0$$ There is only one change of signs in consecutive nonzero coefficients of the quartic polynomial. Thus, there exactly one real positive solution $u^*$ (Decartes criteria) and $L^*=(u^*)^2$ is the limit of $a_n$.

  • Applying Descartes' to $p(-u)=u^4+u-2=0$, you get that there is also exactly one positive solution, which happens to be $u_*=1$. That results in the extra positive solution that you get from the quartic equation in $L$ in your posting.

  • One last comment: the Descartes' criteria applied to $$q(L)=L^3+L^2-3L-4=0$$ also implies that there is only one positive solution to the cubic equation, for there is exactly one change of sign in consecutive nonzero coefficients of $q$.

Mittens
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