why is this sequence convergent Suppose to have two sequence $(a_n)_{n\geq1}$ and $(b_n)_{n\geq1}$ such that
$a_n=\frac{1}{2}(a_{n-1}+b_{n-1})$.
I want to prove that if $b_n\rightarrow0$ then $a_n\rightarrow0$.
The only thing I was able to prove is that $a_n$ is bounded, in fact:
$b_n$ is convergent and so bounded $|b_n|\leq M$. And so $|a_n|\leq |\frac{a_2}{2^{n-1}}+\frac{b_2}{2^{n-2}}+\cdots+\frac{b_{n-1}}{2}|\leq|\frac{a_2}{2^{n-2}}|+M\sum^\infty_{k=1}\frac{1}{2^k}$ and for great $n$ we have $|\frac{a_2}{2^{n-2}}|\leq\varepsilon$
Could you help me to continue (if I'm on the right track), please?
I see that if we prove that $a_n$ has a limit $L$ then necessarily $L=0$ because $L$ must satisfy $L=\frac{1}{2}L$, but I don't know how to prove that it has a limit.
 A: Hint: You have a good start in proving that the sequence $(a_n)$ is bounded. Let's reuse your trick and look at
$$
a_{2n}=\frac{a_n}{2^n}+\frac{b_{2n-1}}2+\frac{b_{2n-2}}4+\cdots+\frac{b_n}{2^n}.
$$
All the numbers $|b_k|<\epsilon$ for $k\ge n$, if $n$ is large enough. The first term $a_n/2^n$ looks like it would be under control as well now that you know $|a_n|$ to be bounded.
A: Assume that $-t\leqslant b_n\leqslant t$ for a given positive $t$ and for every $n\geqslant n_t$. Then $a_n-t\leqslant\frac12(a_{n-1}-t)$ and $a_n+t\geqslant\frac12(a_{n-1}+t)$ for every $n\geqslant n_t$. Thus $\limsup (a_n-t)\leqslant0$ and $\liminf (a_n+t)\geqslant0$. Since this holds for every positive $t$, $a_n\to 0$.
A: One way to do this is to view the recursion equation defining $a_n$ as a, well, recursion equation. It is then a non-homogeneous linear recursion, whose associated homogeneous recursion is very easy to solve. One could then use Lagrange's method of variation of parameters to determine the actual solution, but instead of doing that (we can't, in fact, because we do not know $b_n$) we can use the same idea  to obtain bounds that will prove what you want. Let's do that.
Let $\varepsilon>0$. There exists an $N$ such $|b_n|\leq\varepsilon$ for all $n\geq N$.
Let $a_n=\frac{\alpha_n}{2^n}$ (this is where we use Lagrange's method: the solution for the homogeneous equation is $\frac{\alpha}{2^n}$, with $\alpha$ a constant and Lagrange suggests that we now turn $\alpha$ into a function  $\alpha_n$ of $n$) and suppose that $|b_n|\leq\beta$ for all $n\geq1$. Replacing this in the defining recursion tells us that $$|\alpha_n-\alpha_{n-1}|=2^{n-1}|b_n|$$ for all $n\geq1$. This implies that when $n\geq N$ \begin{align}|\alpha_n-\alpha_0|&\leq|\alpha_n-\alpha_{n-1}|+|\alpha_{n-1}-\alpha_{n-2}|+\cdots+|\alpha_1-\alpha_0| \\ &\leq(2^{n-1}+\cdots+2^N)\varepsilon + (2^{N-1}+\cdots+2^0)\beta\\&\leq (2^n-2^N)\varepsilon+(2^N-1)\beta \\&\leq 2^n \varepsilon+c\end{align} for some positive constant $c$. Then $$|\alpha_n|\leq 2^n\varepsilon+c+|\alpha_0|$$ and $$|a_n|=|\alpha_n/2^n|\leq\varepsilon+\frac{c+|\alpha_0|}{2^n}.$$ This should make it clear that $a_n\to0$.
