I believe that I have managed to show that (if $x\gt 0$) $$\lim_{n\to\infty} \sqrt{n}\cdot{\overbrace{\sin\sin\cdots\sin}^{n\space\text{sines}}(x)}=\sqrt{3}$$ I did this by defining a sequence as $a_0=x$ and the recursion $$a_{n+1}=\sin a_n$$ I then approximated the recursion with the first two nonzero terms of the Maclaurin series for sine, giving me $$\Delta a_n=-\frac{x^3}{6}$$ I then approximated this with a differential equation $$y'=-\frac{y^3}{6}$$ Which I then easily solved... the answer follows from here.

Question: How can this be made more rigorous? I don't know how to justify that my approximations are good enough for th error to vanish under the limit. What theorems are generally used to justify approximations of discrete recursions with differential equations? I think I know how to justify the approximation of some with its Maclaurin series using the Lagrange error bound.

  • $\begingroup$ @Dzoooks It is a finite difference. $$\Delta a_n:=a_{n+1}-a_n$$ $\endgroup$ Commented Jul 30, 2018 at 15:48
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    $\begingroup$ How is this possible? $$|\overbrace{\sin \dots \sin}^{n \ \text{sines}}(x)| \leq 1,$$ so your limit is 0 by the Squeeze Theorem. Should you be dividing by $\sqrt{x}$? $\endgroup$
    – Dzoooks
    Commented Jul 30, 2018 at 15:51
  • $\begingroup$ This is true for $x \in (0, \pi/2): \lim a_n\sqrt n = \sqrt 3$, which could be justified by applying Stolz formula. $\endgroup$
    – xbh
    Commented Jul 30, 2018 at 15:53
  • $\begingroup$ @xbh and Dzoooks: sorry, thanks for pointing that out. The $\sqrt n$ should be multiplied. $\endgroup$ Commented Jul 30, 2018 at 15:56
  • $\begingroup$ @xbh Can you explain why this only works for x in that interval? And if you walked me through the application of Stolz and how to carefully justify this, that would be the kind of answer I am looking for. In particular, I would like to justify the fact that the error is small enough to be negligible as n goes to infinity. $\endgroup$ Commented Jul 30, 2018 at 16:00

4 Answers 4


This may not be the answer you wanted, cuz it may not be generalized.

Claim: $\lim n a_n^2 = 3$.

Proof. $\blacktriangleleft$ Note that not matter which value $x$ is, $a_1 = \sin (x) \in (-1,1)$, so it suffice to consider if $x \in (0, \pi/2)$. Now $a_1 > 0$ and $a_2 < a_1$. By induction it would be easy to verify that $0 < a_n< a_{n-1}$. Hence $a_n \searrow$ and $a_n > 0$. Therefore the limit $A = \lim a_n$ exists. Solve $\sin (A) = A$ in $[0, \pi/2]$ gives that $A=0$, i.e. $a_n = o(1)$.

Now compute the limit: \begin{align*} \lim n a_n^2 &= \lim \frac n {a_n^{-2}} \\ &= \lim \frac 1 {a_{n+1}^{-2} - a_n^{-2}}\quad [\text {Stolz formula}]\\ &= \lim \frac {a_n^2 a_{n+1}^2} {a_n^2 - a_{n+1}^2} \\ &= \lim \frac {a_n^2 \sin^2(a_n)} {(a_n - \sin(a_n) )(a_n + \sin (a_n))} \\ &= \lim \frac {a_n^4} {a_n^3/6 \cdot 2a_n} \quad [\sin(a_n) \sim a_n; a_n - \sin (a_n) \sim a_n^3/6]\\ &= 3. \end{align*} Hence $\lim a_n \sqrt n = \sqrt 3. \blacktriangleright$

  • $\begingroup$ Even if you start with $x \ge \pi/2$, already the next iterate $\sin x$ is in the interval $[0,\pi/2)$, so there's no problem. $\endgroup$ Commented Jul 30, 2018 at 18:15
  • $\begingroup$ @HansLundmark Thanks, I forgot about this. $\endgroup$
    – xbh
    Commented Jul 30, 2018 at 18:30
  • $\begingroup$ Very clever use of Cesaro-Stolz. +1 $\endgroup$
    – Paramanand Singh
    Commented Jul 31, 2018 at 0:43
  • $\begingroup$ @ParamanandSingh Thanks! $\endgroup$
    – xbh
    Commented Jul 31, 2018 at 2:07

I'll assume $0<x<1$, so that your sequence decreases to zero.

Write $$b_n=a_n^2=\sin^2(\sin^{[n-1]}(x))$$ where I'll use $\sin^{[m]}(x)$ for the $m$-fold composite of $\sin$. We'd like to prove $b_n\sim 3/n$. But $b_{n+1}=\phi(b_n)$ where $$\phi(t)=\sin^2\sqrt t=t-\frac{t^2}3+\frac{t^3}{12}+\cdots.$$ Then $$\frac1{b_{n+1}}=\frac1{b_n}\left(1+b_n+O(b_n^2)\right) =\frac1{b_n}+\frac13+O(b_n).$$ As $b_n\to0$ then $1/b_n$ increases by at least $1/4$ eventually, so that $1/b_n<n/4+C$ for some constant $C$, and so $b_n=O(1/n)$. Then$$\frac1{b_n}=\frac n3+O(\ln n)$$ which is good enough.

  • $\begingroup$ Better than mine. Thanks for the elegant analysis! $\endgroup$
    – xbh
    Commented Jul 30, 2018 at 16:19
  • $\begingroup$ This is hard to understand. How did you obtain the expression for $1/b_{n+1}$? $\endgroup$ Commented Jul 30, 2018 at 16:47
  • $\begingroup$ @Frpzzd By writing $1/b_{n+1}=1/\phi(b_n)$ and taking the first few terms of the Laurent series. $\endgroup$ Commented Jul 30, 2018 at 16:49
  • $\begingroup$ @AnginaSeng, you should write $$\frac1{b_{n+1}}=\frac1{b_n}\left(1+\color{red}{\dfrac{b_n}3}+O(b_n^2)\right) =\frac1{b_n}+\frac13+O(b_n)\;.$$Why did you write that $1/b_n$ increases by at least $1/4$ eventually ? $\endgroup$
    – Angelo
    Commented Dec 18, 2023 at 21:30

Since you asked for a reference, here is one that I have enjoyed for many years:

Asymptotic Methods in Analysis by N. G. de Bruijn, available quite inexpensively at Dover and Amazon.

Chapter 8 has an extensive discussion of iterated functions which includes most of the methods in the answers to this question.

  • $\begingroup$ Thank you very much! I will have to add this to my "to read" list. $\endgroup$ Commented Aug 2, 2018 at 19:22

We have

$$ \sqrt n \sin(\cdots \sin x) = a_n\\ \sqrt{n+1}\sin(\sin(\cdots \sin x)) = a_{n+1} $$


$$ \sin\left(\frac{a_n}{\sqrt n}\right) = \frac{a_{n+1}}{\sqrt{n+1}} $$

Analyzing now the iterative procedure

$$ u_{n+1} = \sin u_n $$

with $u_0 = \frac{\pi}{2}$ the sequence

$$ \sqrt k u_k $$

is convergent and has a behavior shown in the attached plot

(in red $\sqrt 3$ and in blue $\sqrt k u_k$)

enter image description here


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