Let $a_1$ = $\sqrt{2}$ and $a_{n+1} = \sqrt{1+\sqrt{a_n}}$ for all natural numbers n.
Show that the sequence $a_n$ converges

I've been doing problems which use the Root/Ratio test. I don't see how I can use that here.
I started with $\sqrt2$ < $2$, and $\sqrt{a_1}$ < $1$ and tried to reason out the series. That doesn't work. How would I show that it converges to some value $L$?

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    $\begingroup$ You seem to be confusing sequence and series. (The title says "series," but you mean sequence. On the other hand, the root/ratio tests apply only to series!) $\endgroup$ – Ted Shifrin Sep 15 '16 at 23:54

The sequence is monotone increasing. Show that it is bounded from above and you are done.

Monotonicity. Proof by induction. We have $$a_1 - a_0 = \sqrt{1 + \sqrt{2}} - \sqrt{2} \approx 0.134 > 0$$ Hence assume the sequence is monotone increasing for $n > 0$. We have $$a_{n + 1} - a_n = \sqrt{1 + \sqrt{a_n}} - \sqrt{1 + \sqrt{a_{n-1}}} = \frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{\sqrt{1 + \sqrt{a_n}} + \sqrt{1 + \sqrt{a_{n-1}}}} = \frac{a_n - a_{n-1}}{\left( \sqrt{1 + \sqrt{a_n}} + \sqrt{1 + \sqrt{a_{n-1}}}\right)\left( \sqrt{a_n} + \sqrt{a_{n-1}}\right)} > 0$$ since by the induction hypothesis $a_n - a_{n-1} > 0$ and the sequence is for sure positive (one can show by induction, that the sequence is bounded below by $1$).

Boundedness. The sequence is bounded above by $2$ (this is not sharp, but its an upper bound). Obviously, $\sqrt{2} \leqslant 2$. Assume, $a_n \leqslant 2$. Then $$a_{n + 1} = \sqrt{1 + \sqrt{a_n}} \leqslant \sqrt{1 + \sqrt{2}} \leqslant \sqrt{4} = 2$$ since the root function is monotone increasing and of course $\sqrt{2} \leqslant 3$.

Hence we can conclude, that the sequence converges.

Limit. If you want to calculate $\lim_{n \to \infty} a_n$ we stick to the following trick. Denote the limit by $L$. Then it must hold $$L = \lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{1 + \sqrt{a_n}} = \sqrt{1 + \sqrt{L}}$$ since the root function is continuous. Hence we have to solve $$L = \sqrt{1 + \sqrt{L}}$$ or equivalently $$L^4 - 2L^2 - L + 1 = 0$$ However this is not so easy to solve. Perhaps, someone has a good idea to solve this, but using Wolfram, we get the two roots $$L_1 \approx 0.524829 \qquad L_2 \approx 1.4902$$ So for sure $$\lim_{n \to \infty} a_n \neq L_1$$ since the sequence is monotone increasing and bounded below by $1$. Thus $$\lim_{n \to \infty} a_n = L_2 \approx 1.4902 $$

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    $\begingroup$ How do you show it's monotone increasing, though? That's a quite important part. $\endgroup$ – Clement C. Sep 15 '16 at 23:45
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    $\begingroup$ @ClementC. Just edited it. $\endgroup$ – TheGeekGreek Sep 15 '16 at 23:59

Consider $f(x)=\sqrt{1+\sqrt x}$ on $\{x>0\}$, $f'(x)={1\over 4}{1\over \sqrt{1+\sqrt x}}{1\over\sqrt x}>0$ thus $f$ is an increasing function. Since $a_{n+1}=f(a_n)$ and $f(a_1)\geq a_1$, we deduce recursively that $(a_n)$ is an increasing sequence.

Now let $g(x)=\sqrt{1+x}$, write $b_1=\sqrt2$ and $b_{n+1}=g(b_n)$. Remark that $g'(x)={1\over 2}{1\over\sqrt{1+x}}>0$. Since $g(a_1)>a_1$. We deduce that $(b_n)$ is an increasing sequence.

Remark that $f(x)\leq g(x)$ if $x>1$. Suppose that $1\leq a_n\leq b_n$, $a_{n+1}=f(a_n)\leq g(a_n)\leq g(b_n)=b_{n+1}$. Since $a_1=b_1$, we deduce that for every integer $n$, $a_n\leq b_n$.

$g(x)=x$ i.e $\sqrt{1+x}=x$ i.e $x^2-x-1=0$. The positive root of this equation is ${{1+\sqrt5}\over 2}>\sqrt2$. Suppose that $a_n\leq b_n\leq {{1+\sqrt5}\over 2}$, $a_{n+1}\leq b_{n+1}=g(b_n)\leq g({{1+\sqrt5}\over 2})={{1+\sqrt5}\over 2}$. Since $a_1=b_1\leq {{1+\sqrt5}\over 2}$, we deduce recursively that $(a_n)$ is an increasing bounded sequence thus converges.

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    $\begingroup$ From $f'(X)>0$ you can not prove $f(X) > X$. Say, $f(X)=X-1$. $\endgroup$ – S. Y Sep 16 '16 at 1:47
  • $\begingroup$ And actually, here this crucially depends on the initial condition. If we had had $a_1=3$, say, one can check the sequence would have been decreasing instead. (It is only increasing ecause $a_1=\sqrt{2}$ is less than the fixed point of the function, which is roughly $1.49$.) $\endgroup$ – Clement C. Sep 16 '16 at 2:28
  • $\begingroup$ who downvote here and why? $\endgroup$ – Tsemo Aristide Sep 16 '16 at 2:32
  • $\begingroup$ I did, for the reason above. Your argument was (and still is as I write this comment) wrong. $\endgroup$ – Clement C. Sep 16 '16 at 2:36
  • $\begingroup$ Why is it wrong, even if $a_1=3, f'>0$. $\endgroup$ – Tsemo Aristide Sep 16 '16 at 2:38

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