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I have $a_k=\frac1{(k+1)^\alpha}$ and $c_k=\frac1{(k+1)^\lambda}$, where $0<\alpha<1$ and $0<\lambda<1$, and we have a infinite sequence $x_k$ with the following evolution equation. $$ x_{k+1}=\left(1-a_{k+1}\right)x_{k}+a_{k+1}c_{k+1}^{2} $$ I have proven that $x_k$ is bounded and obviously positive. How can I know its limit?

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If you call the limit $x$ and substitute $x+\delta$ near $x$ into the evolution equation, the resulting $x_{k+1}$ has to be near $x$ again, in the sense that it goes to zero as $\delta$ goes to zero; else it cannot go to zero as the sequence converges to the limit. That gives you an equation for $x$. – joriki Mar 29 '11 at 5:13
More generally, it is possible completely solve and find $x_n$ in $$x_{n+1}=f(n) x_n+g(n)$$ for arbitrary functions $f(n),g(n)$. The solution then depends on certain sums and products of the $f(n)$ and $g(n)$. The solution "looks" very similar to that given by solving the first order linear ODE $$y^' +f(t)y=g(t).$$ Indeed, wherever there is an integral, replace with a sum, and wherever you find the exponential of an integral, replace with a product. – Eric Naslund Mar 29 '11 at 5:19
up vote 2 down vote accepted

Let us show that $x_k\to0$.

For every positive $u$, there exists a finite integer $k(u)$ such that $c_{k+1}^2\le u$ for every $k\ge k(u)$, hence $x_{k+1}\le(1-a_{k+1})x_k+a_{k+1}u$.

-- If there exists $k\ge k(u)$ such that $x_k\le u$, then $x_i\le u$ for every $i\ge k$, hence $\limsup x_i\le u$.

-- Otherwise, $x_k\ge u$ for every $k\ge k(u)$ and furthermore $(x_{k+1}-u)\le(1-a_{k+1})(x_k-u)$. Hence $(x_k-u)\le b(k,k(u))(x_{k(u)}-u)$, with $$ b(k,k(u))=\prod_{i=k(u)+1}^k(1-a_i). $$ Now, the series $\displaystyle\sum_ka_k$ diverges hence $b(k,k(u))\to0$ when $k\to+\infty$, and $\limsup x_k-u\le0$.

In both cases, $\limsup x_i\le u$. This holds for every positive $u$, hence $x_k\to0$.

The proof uses only that $c_k\to0$, $a_k\in[0,1]$ and $\displaystyle\sum_ka_k$ diverges.

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Great, it is shown that $x_k$ can be made arbitrarily small, although we don't know the convergence trajectory exactly. – user8827 Mar 30 '11 at 3:19
Yes, this is a good remark. – Did Mar 30 '11 at 11:05

As pointed out in the comments, a general solution to $x_{n+1}=f_n x_n + g_n$ is possible. Define $F_0=1$ and $F_{n+1} = \prod_{i=0}^{n} f_n^{-1} = f_n^{-1} F_{n}$ for $n\ge 0$. Then $f_n = F_{n} / F_{n+1}$, and the recurrence relation becomes $$ F_{n+1} x_{n+1} = F_{n} x_{n} + F_{n+1} g_n. $$ This has the solution $F_{n+1} x_{n+1} = F_{0} x_{0} + \sum_{i=0}^{n} F_{i+1} g_{i}$, or $$ x_{n+1} = \frac{F_0}{F_{n+1}} x_{0} + \sum_{i=0}^{n} \frac{F_{i+1}}{F_{n+1}} g_{i} = x_{0}\prod_{i=0}^{n} f_{i} + \sum_{i=0}^{n} g_{i} \prod_{j=i+1}^{n} f_{j}. $$ Note that the final product is empty when $i=n$, and has the value $1$ in that case. In the specific problem given, $f_{i}=1-(i+2)^{-\alpha}$ and $g_{i} = (i+2)^{-\alpha-2\lambda}$ (up to possible $\pm 1$ mistakes in either the problem or my solution). The partial products $\prod_{i=0}^{n} f_{i}$ diverge to zero, so the first term vanishes for large $n$, and the limit of the sequence is independent of $x_0$. The result is that $$ \lim_{n\rightarrow\infty}x_{n} = \lim_{n\rightarrow\infty}\left[\sum_{i=0}^{n} (i+2)^{-\alpha-2\lambda} \prod_{j=i+1}^{n} \left(1-(j+2)^{-\alpha}\right)\right], $$ provided that the limit exists (which doesn't seem obvious).

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As an additional comment, the limit is $0$ when $\alpha + 2\lambda > 1$; the more difficult case is when the sum of the $g_i$ diverges to infinity while the product of the $f_i$ diverges to $0$. – mjqxxxx Mar 29 '11 at 15:22
Quite a general solution to the problem. It would be better if a more specific upper bound of $x_k$ could be found, e.g. $x_k\le f(k)$ and We can derive the final limit by further proof on $f(k)\to 0$ as $k\to\infty$. – user8827 Mar 30 '11 at 4:05

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