Let $\{a_n\}_{n=1}^\infty$ be a sequence of real numbers and define the sequential average $\bar{a}_n=\frac{1}{n}\sum_{i=1}^n a_i$ and a demeaned sequence $$ b_n=a_n-\bar{a}_{n-1},\quad n\geq 1, $$ where $\bar{a}_0=0$.
If we know that $b_n\to 0$ as $n\to\infty$, is it true that $\{a_n\}$ is bounded? If not, is there a condition on the convergence rate of $b_n$ so that $a_n$ must be bounded?
While writing this answer it appears to me that I discovered some sort of a tautology, basically because $a_n$ are uniquely determined by $b_n$. Specifically, by simple algebra we have $\bar{a}_n - \bar{a}_{n-1} = \frac{b_n}{n}$. From this it follows that $\bar{a}_n=\sum_{k=1}^n \frac{b_k}{k}$ and $a_n = b_n + \sum_{k=1}^{n-1} \frac{b_k}{k}$.
If $\sum_{k = 1}^n \frac{b_k}{k}$ and $b_n$ are bounded sequences, then $a_n$ is bounded as a sum of bounded sequences. Conversly, if $a_n$ are bounded then $\bar{a}_n$ are bounded since they are averages of uniformly bounded numbers, hence $b_n$ is bounded as a difference of bounded sequences.
So, $a_n$ is bounded if and only if $b_n$ is bounded and $\sum_{k=1}^n \frac{b_k}{k}$ is bounded. In particular, there exists $b_k \to 0$ such that $a_n$ is unbounded, for example $b_k = \frac{1}{\log k}$. If you only know $|b_k| \le c_k$ (the decay condition you mentioned) then what you need is $c_k$ being bounded and $\sum_{k=1}^\infty \frac{c_k}{k} < \infty$.
