Consider the sequence defined as

$x_1 = 1$

$x_{n+1} = \sin x_n$

I think I was able to show that the sequence $\sqrt{n} x_{n}$ converges to $\sqrt{3}$ by a tedious elementary method which I wasn't too happy about.

(I think I did this by showing that $\sqrt{\frac{3}{n+1}} < x_{n} < \sqrt{\frac{3}{n}}$, don't remember exactly)

This looks like it should be a standard problem.

Does anyone know a simple (and preferably elementary) proof for the fact that the sequence $\sqrt{n}x_{n}$ converges to $\sqrt{3}$?


3 Answers 3


Before getting into the details, let me say: The ideas I'm talking about, including this exact example, can be found in chapter 8 of Asymptotic Methods in Analysis (second edition), by N. G. de Bruijn. This is a really superb book, and I recommend it to anyone who wants to learn how to approximate quantities in "calculus-like" settings. (If you want to do approximation in combinatorial settings, I recommend Chapter 9 of Concrete Mathematics.)

Also, this isn't just about $\sin$. Let $f$ be a function with $f(0)=0$ and $0 \leq f(u) < u$ for $u$ in $(0,c]$ then the sequence $x_n:=f(f(f(\cdots f(c)\cdots)$ approaches $0$. If $f(u)=u-a u^{k+1} + O(u^{k+2})$ (with $a>0$) then $x_n \approx \alpha n^{-1/k}$ and you can prove that by the same methods here.

Having said that, the answer to your question. On $[0,1]$, we have $$\sin x=x-x^3/6+O(x^5).$$ Setting $y_n=1/x_n^2$, we have $$1/x_{n+1}^2 = x_n^{-2} \left(1-x_n^2/6+O(x_n^4) \right)^{-2} = 1/x_n^2 + 1/3 + O(x_n^2)$$ so $$y_{n+1} = y_n + 1/3 + O(y_n^{-1}).$$

We see that $$y_n = \frac{n}{3} + O\left( \sum_{k=1}^n y_k^{-1} \right)$$ and $$\frac{1}{n}y_n = \frac{1}{3} + \frac{1}{n} O\left( \sum_{k=1}^n y_k^{-1} \right)$$ Since we already know that $x_n \to 0$, we know that $y_n^{-1} \to 0$, so the average goes to zero and we get $\lim_{n \to \infty} y_n/n=1/3$. Transforming back to $\sqrt{n} x_n$ now follows by the continuity of $1/\sqrt{t}$.

  • 33
    $\begingroup$ PS This is a good example of why I find the O() notation insanely more useful than limits. $\endgroup$ Commented Aug 24, 2010 at 18:12
  • $\begingroup$ Well done! Brute force wins! This is a very straightforward proof without the need to appeal to a nested set of theorems, each of which requires proof. $\endgroup$
    – Mark Viola
    Commented Mar 5, 2015 at 19:19

This problem can be found in Kaczor, Nowak: Problems in Mathematical Analysis I, Real Numbers, Sequences and Series. I'll copy their solution here.

Problem 2.5.22, p.50, a solution is given on p.215.

Problem 2.5.22. The sequence $(a_n)$ is defined inductively as follows: $$0<a_1<\pi \qquad a_{n+1}=\sin a_n \text{ for }n\ge 1$$ Prove that $\lim\limits_{n\to\infty} \sqrt n a_n = \sqrt3$.

Solution: It is easy to see that the sequence $(a_n)$ is monotonically decreasing to zero. Moreover, an application of I'Hospital's rule gives $$\lim\limits_{x\to 0}\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac13.$$ Therefore $$\lim\limits_{n\to\infty}\left(\frac1{a_{n+1}^2}-\frac1{a_n^2}\right)=\frac13$$ Now, by the result in Problem 2.3.14, $\lim\limits_{n\to\infty} na_n^2 = 3$.

Problem 2.3.14, p.38, a solution is given on p.184.

Problem 2.3.14. Prove that if $(a_n)$ is a sequence for which $$\lim\limits_{n\to\infty}(a_{n+1}-a_n)=a$$ then $$\lim\limits_{n\to\infty}\frac{a_n}n=a.$$

Solution: In Stolz theorem we set $x_{n}=a_{n+1}$ and $y_n=n$.

Formulation of Stolz theorem in this book is the following

Let $(x_n)$, $(y_n)$ be two sequences that satisfy the conditions:

  • $(y_n)$ strictly increases to $+\infty$,
  • $$\lim\limits_{n\to\infty} \frac{x_n-x_{n-1}}{y_n-y_{n-1}}=g.$$

Then $$\lim\limits_{n\to\infty} \frac{x_n}{y_n}=g.$$

For Stolz-Cesaro theorem see also this question: Stolz-Cesàro Theorem

Perhaps it is also worth mentioning that there are two equivalent forms of Stolz-Cesaro theorem: see e.g. this answer.

  • 5
    $\begingroup$ Another source: Polya-Szego, Problem and Theorems in analysis I, Spronger-Verlag, problems 173-174, p.38 $\endgroup$ Commented May 25, 2012 at 16:26
  • $\begingroup$ .........Thanks! $\endgroup$
    – Aryabhata
    Commented May 25, 2012 at 19:17
  • $\begingroup$ Now I noticed that this question also has an answer based on Stolz-Cesaro: Need help solving Recursive series. $\endgroup$ Commented Jun 17, 2012 at 6:07
  • $\begingroup$ I wish to you all the best :-) I not forgot. Thank you very much .... $\endgroup$
    – Sebastiano
    Commented Dec 9, 2020 at 22:53

If your sequence is of the form $$x_{n+1} = f(x_n)$$ with $$f(x) = x( 1 - c x^{\alpha} + \textrm{h. o. t} )$$ then notice ( e.g. ) that $$\frac{1}{f(x)^{\alpha}} = \frac{1}{x^{\alpha}} + \alpha c+ \textrm{h. o. t.}$$ and so $$\frac{1}{x_{n+1}^{\alpha}} - \frac{1}{x_n^{\alpha}} \to \alpha c$$ and with Cesaro-Stolz we get $$\frac{1}{n x_n^{\alpha} } \to \alpha c$$

In the particular case $f(x) = \sin x= x( 1 - \frac{x^2}{6} + \cdots)$, we get $$\frac{1}{n x_n^2} \to 2 \cdot \frac{1}{6} = \frac{1}{3}$$

$\bf{Added}$ We can get effective estimates for the sequence

$x_{n+1} = \sin x_n$ as follows

Consider the Taylor expansion

$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} - 1}{x^2} = \frac{1}{5} + \frac{2 x^2}{63} + \cdots$$

where all the remaining coefficients are positive. We conclude that for $0<|x|< \pi$ we get

$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} > \frac{1}{5}$$

and for $0 < |x|< 1$ we have

$$\frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} < \frac{\frac{3}{\sin^2 x} - \frac{3}{x^2} -1}{x^2} \ _{x=1}=0.2368\ldots < \frac{1}{4}$$

Therefore we have for $0< |x|\le 1$

$$\frac{3}{x^2} + 1 + \frac{x^2}{5}<\frac{3}{\sin^2 x} < \frac{3}{x^2} + 1 + \frac{x^2}{4}$$

Now for the sequence $y_n= \frac{3}{x_n^2}$ we get the inequalities

$$y_n+ 1 + \frac{3}{5 y_n} < y_{n+1} < y_n + 1 + \frac{3}{4 y_n}$$

and we can use this to get estimates for $y_n$ and so $x_n$.


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