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Given a sequence $x_1=\frac{1}{2}$, $x_{n+1}=x_n-x_n^2$. It's easy to see that it limits to $0$.

The question is: is there exists an $\alpha$ such, that $\lim\limits_{n\to\infty}n^\alpha x_n\neq0$.

I tried to find explicit formula for $x_n$, but did not succeed. I stucked at that point: if $a_1=(\frac{1}{4})^2$, $a_n=(a_{n-1}+\frac{1}{4})^2$, then $x_n=-a_{n-2}+\frac{1}{4}$ for $n\ge3$.

Thanks for any ideas and help.

Makarchi
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2 Answers2

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Let $y_n = \frac{1}{x_n} - n$, we have

$$y_{n+1} = \frac{1}{x_{n+1}}-(n+1) = \frac{1}{x_n} + \frac{1}{1-x_n} - (n+1) = y_n + \frac{x_n}{1-x_n}$$ This implies for $n > 1$,

$$y_n = y_1 + \sum_{k=1}^{n-1} \frac{x_k}{1-x_k} \quad\iff\quad \frac{1}{x_n} = n + 1 + \sum_{k=1}^{n-1}\frac{x_k}{1-x_k} \tag{*1}$$ Since $0 \le x_k < 1$, this immediately give us an upper bound of $x_n$:

$$\frac{1}{x_n} \ge n+1 \quad\iff\quad x_n \le \frac{1}{n+1}\tag{*2}$$

Substitute $(*2)$ in $(*1)$ and notice the map $x \mapsto \frac{x}{1-x}$ is an increasing function for $x \in [0,1)$, we obtain a lower bound for $x_n$:

$$\begin{align} & \frac{1}{x_n} \le n+1 + \sum_{k=1}^{n-1}\frac{\frac{1}{k+1}}{1 - \frac{1}{k+1}} = n + 1 + \sum_{k=1}^{n-1}\frac{1}{k} \le n + 1 + \log n + \gamma\\ \implies & x_n \ge \frac{1}{n + 1 + \log n + \gamma}\tag{*3} \end{align}$$ where $\gamma$ is the Euler–Mascheroni constant.

Combine these two bounds, it is clear we can take $\alpha = 1$ and $\displaystyle\lim_{n\to\infty} n^\alpha x_n = 1$.

achille hui
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  • Everything great! Thanks =) At first seemed a little mystery why we consider $y_n=\frac{1}{x_n}-n$, but now I get it. (We are looking for a bound $C_1(n+c_1)^\alpha\ge x_n\ge C_2(n+c_2)^\alpha$. ) – Makarchi Sep 27 '13 at 10:40
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    @Makarchi The idea is for large $n$, one should be able to approximate $x_n$ by $X(n)$ for some smooth function $X$. The equation $x_{n+1} = x_n - x_n^2$ then becomes something like $$\frac{dX(n)}{dn} = -X(n)^2 \quad\iff\quad \frac{d}{dn} \left(\frac{1}{X(n)} - n \right) = 0$$ That prompts me to study the behavior of $y_n$. – achille hui Sep 27 '13 at 10:51
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    Nice answer and an especially nice comment. – Antonio Vargas Sep 28 '13 at 12:55
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There is no formula for the general term of the recurrence $\,x_{n+1}=x_n-x_n^2\,$ however, there is some specific results about the sequence with $\,x_0=1/2.\,$ The denominator of $\,x_n\,$ is $\,2^{2^n}.\,$ OEIS sequence A076628 is the sequence of numerators of $\,x_n\,$ where $\,x_0=\frac12,\, x_1=\frac14,\, x_2=\frac3{16},\,x_3=\frac{39}{256},\, x_4=\frac{8463}{65536},\,\dots.$

In 1999 I found the result (where the constant $c$ depends on $x_0$) $$x_n =1/( n + c + \log(n + c-1/2 + \log( \\ n + c-17/24 + \log(n + c-919/1152 + \dots )))). \tag{1}$$ With $y_n:=1/x_n$ and $L_n:=\log(n)$, then $y_{n+1}=y_n^2/(y_n-1).\;$ The general result is $$ y_n = n + (c_0+c_1L_n) + (c_2+c_3L_n)/n + \\(c_4+c_5L_n+c_6L_n^2)/n^2 + \dots \tag{2}$$ for some constants $c_0,c_1,c_2,\dots$ which depend on $x_0$. Now first replace $n$ with $n+1$ in $y_n$ to get $$ y_{n+1}\!=\!n\!+\!(1\!+\!c_0\!+\!c_1L_n)\!+\!(c_1\!+\!c_2\!+\!c_3L_n)/n \!+\!(c_4\!+\!c_3\!-\!c_2\!-\!c_1/2\!+\!(c_5\!-\!c_3)L_n\!+\!c_6L_n^2)/n^2 \!+\!\dots \tag{3}$$ while using the recursion gives $$y_{n+1}\!=\!n\!+\!(1\!+\!c_0\!+\!c_1L_n)\!+\!(1\!+\!c_2\!+\!c_3L_n)/n \!+\!(c_4\!-\!c_0\!+\!1\!+\!(c_5\!-\!c_1)L_n\!+\!c_6L_n^2)/n^2 \!+\!\dots \tag{4}$$ and equating the two expressions for $y_{n+1}$ we must have $$c_1=c_3=1,c_2=c_0-1/2,c_5=3/2-c_0,c_6=-1/2,c_4=(-5+9c_0-3c_0^2)/6. \tag{5}$$ Making the substitutions for the constants we get $$y_n\!=\!n\!+\!(c_0\!+\!L_n)\!+\!(c_0\!-\!1/2\!+\!L_n)/n\! +\!((-5\!+\!9c_0\!-\!3c_0^2)/6+(3/2-c_0)L_n\!-\!1/2L_n^2)/n^2+\dots \tag{6}$$ and $$x_n=1/n+(-c_0-\log n)/n^2+(c_0^2-c_0+1/2+ \\ (2c_0-1)\log n+(\log n)^2)/n^3+\dots \tag{7}$$ which still depends on $c_0$ which comes from $x_0$.

Somos
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