How do I know the limit of this infinite sequence 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?
 A: 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).
A: 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.
