Give a precise meaning to evaluate the following: $$\large{\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\dotsb}}}}}$$

Since I think it has a recursive structure (does it?), I reduce the equation to

$$ p=\sqrt{1+p} $$ $$ p^2=1+p $$ $$ p^2-p-1=0 $$ $$ p=\frac{1\pm\sqrt{5}}{2} $$

Did I do this right?


marked as duplicate by Semsem, Macavity, Ian Coley, user63181, Claude Leibovici Mar 15 '14 at 7:25

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    $\begingroup$ (1) You first must prove the sequence $$\left\{\sqrt1\,,\,\sqrt{1+\sqrt1}\,,\,\sqrt{1+\sqrt{1+\sqrt1}}\,\ldots\right\}$$ converges, then (2) What you did is correct (arithmetic of limits) $\endgroup$ – DonAntonio Mar 14 '14 at 19:22
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    $\begingroup$ @DonAntonio but his work shows $\rho = \frac{1\pm\sqrt5}2$, wich is erroneous. $\endgroup$ – AlexR Mar 14 '14 at 19:29
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    $\begingroup$ @DonAntonio: But you didn't raise this issue in your original comment, so AlexR was right to call you on it. The OP's solution (with $\pm$) is wrong. $\endgroup$ – TonyK Mar 14 '14 at 19:36
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    $\begingroup$ @TonyK, why would I raise something I think is obvious or almost?! In fact, the "hardest" work here is to prove the sequence converges. That it is a positive one and that its limit is non-negative is the easier, imo, part. OTOH, what the OP write is only $\;p^2-p-1=0\implies p=\frac{1\pm\sqrt5}2\;$ and, hopefully, he hasn't yet concluded about the limit. I can't tell, that's for him to say. $\endgroup$ – DonAntonio Mar 14 '14 at 22:59
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    $\begingroup$ Perhaps you were jumping up and down all the place and thought it was all the others jumping, @TonyK. That happens a lot to many. I was trying to explain to you how some comment to an apparently serious student doesn't have to cover all the apparently obvious details, expecting the serious student will write back asking anything that wasn't clear to him. My try to explain you this was obviusly in vane... $\endgroup$ – DonAntonio Mar 14 '14 at 23:58

The problem asks us to assign a precise meaning to the expression.

Let $a_0=1$, and for every $n\ge 0$, let $$a_{n+1}=\sqrt{1+a_n}.$$ The precise meaning of the expression is $$\rho=\lim_{n\to\infty}a_n.$$

Remark: The limit exists, and a version of your argument shows that the limit is indeed $\frac{1+\sqrt{5}}{2}$.

Here is another example of a similar problem. Assign a precise meaning to $$\rho=1+2+4+8+\cdots.$$

We could (?) say $\rho=1+2\rho$ and therefore (??) $\rho=-1$. It is fairly unlikely (though not impossible) that we would really want to say that $1+2+4+\cdots$ means $-1$.

  • $\begingroup$ Same numbering as me ^^ $\endgroup$ – AlexR Mar 14 '14 at 19:23
  • $\begingroup$ Doesn't everybody start counting at $0$? $\endgroup$ – André Nicolas Mar 14 '14 at 19:31
  • $\begingroup$ I actually thought whether I would start at $0$ or at $1$ and decided to start at $0$ because the starting value doesn't matter as long as $\rho_0 \ge -1$ (chosing any $\rho_0$ will just affect the rate of convergence as long as we stay in the reals) $\endgroup$ – AlexR Mar 14 '14 at 19:33
  • $\begingroup$ I must comment on the last sentence: what can $1+2+4+...$ mean other than that? (actually I did not know of that result, but have seen arguments for $1+2+3+...=\frac{-1}{12}$, so...) Two opposite results cannot both be true. Maybe $\displaystyle\lim_{n\to\infty}\sum_{i=0}^n2^n=\infty$, but that is another question. +1 nonetheless. $\endgroup$ – JMCF125 Mar 14 '14 at 19:49
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    $\begingroup$ Thank you, it is $-1$. Numerical meaning is assigned to infinite series, products, iterated operations of various kinds through limits. Given that the question has the tag calculus, it is likely that an answer of the flavour I gave is expected by whoever assigned it. $\endgroup$ – André Nicolas Mar 14 '14 at 20:17

To "give a precise meaning to" is quite broad. A general approach would be to set $$\rho_0 = 1; \qquad \rho_{n+1} = \sqrt{1+ \rho_n}$$ Then to show that $(\rho_n)_n$ is convergent (i.e. cauchy in $\mathbb R$) and to define $\rho$ as the limit, using the completeness of $\mathbb R$. $$\rho = \lim_{n\to\infty} \rho_n$$ You can then prove that $$\rho = \frac{1+\sqrt5}2$$

Note that $\rho\neq\frac{1-\sqrt5}2$ simply by showing that each $\rho_n \ge 1>\frac12 >\frac{1-\sqrt5}2$ Also, the choice of $\rho_0$ is arbitrary as long as $\rho_0 \ge -1$. It will only affect the rate of convergence. The closer $\rho_0$ is to $\frac{1+\sqrt5}2$, the faster the sequence will converge.

  • $\begingroup$ You could also say that $\sqrt{1 + \sqrt{1 + \sqrt{1+...}}}$ means that you can choose $\rho_0$ freely (this is just one possible interpretation of '...'), and then set $ \rho = \lim_{n \rightarrow \infty}{\rho_n}$. Then the solution would be $\rho = \frac{1-\sqrt{5}}{2}$ iff $\rho_0 = \frac{1-\sqrt{5}}{2}$, and $\rho = \frac{1+\sqrt{5}}{2}$ otherwise (if I'm not mistaken: I believe I remember this from a similar exercise and I didn't actually do the math again to check it). $\endgroup$ – Ruben Mar 15 '14 at 1:59

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