TITLE: How do I know the limit of this infinite sequence
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.