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.