why is this sequence convergent

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.

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 $t$, $a_n\to 0$.

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 $a_n\to0$.