Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the sequence $\{a_{n}\}$, with $n\ge1$ and $a>0$, defined as:


I'm trying to prove here 2 things: a). the sequence is convergent; b). the sequence's limit when n goes to $\infty$. I may suppose that there must be a proof for this general case. I saw this problem with the case $a=2$ (where it was required to prove only the convergence), but this is just a particular case. The generalization seems to be much more interesting.

share|cite|improve this question
In the special case $a = 4$, the limit of the sequence is $2$. This is because $2^x + 1 = \sqrt{4^x + \sqrt{4^{x+1} + \sqrt{4^{x+2} + \cdots }}}$ whenever the radical converges, which in turn can be verified heuristically with the functional equation $f(x)^2 = 4^x + f(x + 1)$ satisfied by $f(x) = 2^x + 1$. However, the argument is specific to $a = 4$. – Nick Strehlke Jul 14 '12 at 23:12
up vote 6 down vote accepted

Here is a full answer to part (a) and a partial answer to part (b). Call $(a_n(a))_{n\geqslant1}$ the sequence when the value of the parameter is $a$.

One has $a_0(1)=1$ and $a_{n+1}(1)=u(a_n(1))$ for every $n\geqslant0$ with $u(x)=\sqrt{1+x}$. Hence the usual technique shows that the sequence $(a_n(1))_{n\geqslant0}$ is increasing to $a_\infty(1)=\alpha$ where $\alpha$ solves the equation $\alpha=u(\alpha)$, that is, $\color{red}{\alpha=\frac12(1+\sqrt5)}$.

When $a\lt1$, $a_n(a)\leqslant a_n(1)$ and $(a_n(a))_{n\geqslant0}$ is increasing hence $(a_n(a))_{n\geqslant0}$ converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{a}}\lt a_\infty(a)\leqslant \alpha}$.

When $a\gt1$, $\sqrt{1+\sqrt{aa_{n-1}(1)}}\leqslant a_n(a)\leqslant\sqrt{1+\sqrt{a}a_{n-1}(1)}$ and $(a_n(a))_{n\geqslant0}$ is increasing hence it converges to a finite limit $a_\infty(a)$ with $\color{red}{\sqrt{1+\sqrt{\alpha a}}\lt a_\infty(a)\leqslant\sqrt{1+\alpha\sqrt{a}}}$.

To show the upper bound on $a_n(a)$, one carries over every power of $a$ to the left until it reaches the position of $\sqrt{a}$. Crossing a square root sign halves the exponent and $a\gt1$ hence the power of $a$ which just crossed a square root sign is smaller than the preceding one. For example, the first step of the proof uses $a^{n/2}\leqslant a^{n-1}$ to deduce $$ \sqrt{a^{n-1}+\sqrt{a^n}}=\sqrt{a^{n-1}+a^{n/2}\sqrt1}\leqslant\sqrt{a^{n-1}(1+\sqrt1)}=a^{(n-1)/2}\sqrt{1+\sqrt1}, $$ the second step uses $a^{(n-1)/2}\leqslant a^{n-2}$, and so on, until $a^{3/2}\leqslant a^2$ and $a^{2/2}\leqslant a$. A similar reasoning yields the lower bound.

Finally, the map $a\mapsto a_\infty(a)$ is nondecreasing from $\color{red}{a_\infty(0)=1}$ to $\color{red}{a_\infty(+\infty)=+\infty}$.

share|cite|improve this answer

The Herschfeld's Convergence Theorem may help (the more general subject is 'Nested Radical').

Herschfeld's paper "On Infinite Radicals" is available here.

share|cite|improve this answer
thanks for information (+1) – user 1618033 Jul 14 '12 at 20:26
very interesting and helpful that paper! – user 1618033 Jul 14 '12 at 20:32
@Chris'sister: I am glad it helped, – Raymond Manzoni Jul 14 '12 at 20:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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