Let a sequence, $\{x_n\}$ such that: $x_{n+1}=x_n-x_n^3$ and $0<x_1<1$.
1) Prove $\mathop {\lim }\limits_{n \to \infty } {x_n} = 0$
2) Calculate $\mathop {\lim }\limits_{n \to \infty } n{x_n}^2$

So, section (1) is very easy. I didn't really bother to write it down - just show the sequence is monotonically decreasing and bounded bellow by zero.

Section (2) is the real fun, I do familiar with the Lemma says: "If $a_n$ limit is $0$ and $b_n$ is bounded then the limit of $a_nb_n$ is also zero" - But I don't think it can work here.

I tried separating the limit using limits-arithmetic into two limits, but then I got:
$$\mathop {\lim }\limits_{n \to \infty } n{x_n} \cdot \mathop {\lim }\limits_{n \to \infty } {x_n}$$

  • $\begingroup$ is it $n \to \infty$ $\endgroup$ – Suraj M S Jan 25 '14 at 18:34
  • $\begingroup$ it's a typo. will be corrected. Thanks $\endgroup$ – SuperStamp Jan 25 '14 at 18:40
  1. $\lim_{x \to \infty } {x_n}=0$. Because the sequence is monotone and bounded by $0$ and $1$. Last statement is proved by mathematical induction. Passing to the limit in the recurrence relation is obtained limit $0$. Dividing the recurrence relation by $x_n$ and then passing to the limit we find $\lim_{x \to \infty}\frac{x_{n+1}}{{x_n}} = 1$

  2. Applied Cesaro-Stolz's Lemma:

$$\lim_{x \to \infty } n{x_n}^2 = \lim_{x \to \infty }\frac{n}{(\frac{1}{x_n})^2}=\lim_{x \to \infty }\frac{n+1-n}{(\frac{1}{x_{n+1}})^2-(\frac{1}{x_n})^2}= \lim_{x \to \infty }\frac{(x_n)^2(x_{n+1})^2}{(x_n)^2-(x_{n+1})^2} =$$$$=\lim_{x \to \infty }\frac{(x_n)^4(1-(x_{n})^2)^2}{(x_n)^4(1+\frac{x_{n+1}}{x_n})} =\frac{1}{2}$$

  • $\begingroup$ How can you assume that ${nx_n^2} = {n \over \left({{1 \over x_n}}\right)^2}$? $\endgroup$ – SuperStamp Jan 25 '14 at 19:03
  • 1
    $\begingroup$ It is just a rewriting: $a^2 = \frac{1}{\left(\frac{1}{a}\right)^2}$ $\endgroup$ – Clement C. Jan 25 '14 at 19:07
  • $\begingroup$ 1. is not justified and in 2. it is assumed that $\lim\limits_{k\to\infty}\frac{x_{k+1}}{x_k}=1$, but that is not justified. $\endgroup$ – robjohn Jan 26 '14 at 15:12
  • $\begingroup$ also $1 + \frac{1}{n}$ is monotically decreasing and bounded below by $0$, but does not go to $0$ $\endgroup$ – Ant Jan 26 '14 at 20:26
  • $\begingroup$ Passing to the limit in the recurrence relation is obtained limit $0$. $\endgroup$ – medicu Jan 26 '14 at 20:29

Suppose that $0\lt x_k\lt1$, then $0\lt x_k^2\lt1$ and since $x_{k+1}=x_k(1-x_k^2)$, we know that $0\lt x_{k+1}\lt1$. Furthermore, since $x_k^3\gt0$, we have that $x_{k+1}=x_k-x_k^3\lt x_k$.

Thus, $x_k$ is decreasing and bounded below by $0$. Therefore, $\lim\limits_{k\to\infty}x_k$ exists and we have $$ \lim_{k\to\infty}x_k=\lim_{k\to\infty}x_{k+1}=\lim_{k\to\infty}x_k-\left(\lim_{k\to\infty}x_k\right)^3\tag{1} $$ Equation $(1)$ implies that $\lim\limits_{k\to\infty}x_k=0$.

Since $x_{k+1}=x_k-x_k^3$, dividing by $x_k$ and taking limits, we get that $$ \begin{align} \lim_{k\to\infty}\frac{x_{k+1}}{x_k} &=\lim_{k\to\infty}\left(1-x_k^2\right)\\ &=1\tag{2} \end{align} $$ Next notice that $$ \begin{align} \frac1{x_{k+1}^2}-\frac1{x_k^2} &=\frac{x_k^2-x_{k+1}^2}{x_k^2x_{k+1}^2}\\ &=\frac{(x_k-x_{k+1})(x_k+x_{k+1})}{x_k^2x_{k+1}^2}\\ &=\frac{x_k^3(x_k+x_{k+1})}{x_k^2x_{k+1}^2}\\ &=\frac{x_k^2}{x_{k+1}^2}+\frac{x_k}{x_{k+1}}\tag{3} \end{align} $$ Combining $(2)$ and $(3)$ yields $$ \lim_{k\to\infty}\frac1{x_{k+1}^2}-\frac1{x_k^2}=2\tag{4} $$ Apply $\displaystyle\lim_{k\to\infty}a_k=b\implies\lim_{n\to\infty}\frac1n\sum_{k=1}^na_k=b$ to $(4)$ to get $$ \lim_{n\to\infty}\frac1{nx_{n+1}^2}=2\tag{5} $$ which leads to $$ \lim_{n\to\infty}nx_n^2=\frac12\tag{6} $$

Note that the justification for $(5)$ is saying that if a sequence converges, then its Cesaro means converge to the same limit.


For what it's worth:

Since $x_n > 0$ for all $n$, one can safely divide both sides of the recurrence relation by $x_n$ to get that $$ \frac{x_{n+1}}{x_n} = 1 - x_n^2 $$ and thus $$ n x_n^2 = n\left( 1 - \frac{x_{n+1}}{x_n} \right) $$

Now, since $x_n\searrow 0$, $x_n^3=o(x_n)$ and $x_{n+1}\operatorname*{\sim}_{n\to\infty}x_n$; and in particular $\frac{x_{n+1}}{x_n} \xrightarrow[n\to\infty]{} 1$. Suppose you can get a second order expansion of the form $\frac{x_{n+1}}{x_n}\operatorname*{=}_{n\to\infty} 1 + \varepsilon_n + o(\varepsilon_n)$ for some "simple" $\varepsilon_n$ (I would bet on $Cn^{-\alpha}$ for some $\alpha > 0$). Then, $$ n x_n^2 \operatorname*{\sim}_{n\to\infty}n\varepsilon _n $$ and depending on $\varepsilon_n$, that'd give you the limit...

  • $\begingroup$ I'm not that familiar with o-notation, tough I should $\endgroup$ – SuperStamp Jan 25 '14 at 19:09

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