Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\sqrt{3}$?

share|improve this question
    
Please pick a more meaningful title... –  KennyTM Aug 24 '10 at 17:50
    
@KennyTM: Done. PLease let me know if that can be improved. Thanks. –  Aryabhata Aug 24 '10 at 17:55
    
Looks good. :) –  KennyTM Aug 24 '10 at 17:57

2 Answers 2

up vote 43 down vote accepted

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 $1/\sqrt{t}$.

share|improve this answer
21  
PS This is a good example of why I find the O() notation insanely more useful than limits. –  David Speyer Aug 24 '10 at 18:12
    
Very nice! Thanks... –  Aryabhata Aug 24 '10 at 18:56
    
Second paragraph says .. $0\le f(u)\le u$ then .. $f^{(n)}(c) \to 0$. But what if $f(u)=u$? –  Maesumi Jan 11 '13 at 1:04
    
@Maesumi Inequalities fixed, thanks. –  David Speyer Jan 11 '13 at 14:54

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.

share|improve this answer
3  
Another source: Polya-Szego, Problem and Theorems in analysis I, Spronger-Verlag, problems 173-174, p.38 –  Unoqualunque May 25 '12 at 16:26
    
.........Thanks! –  Aryabhata May 25 '12 at 19:17
    
Now I noticed that this question also has an answer based on Stolz-Cesaro: Need help solving Recursive series. –  Martin Sleziak Jun 17 '12 at 6:07

Your Answer

 
discard

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.