To put simply:
When $\bar{a}_n$ is the average of $a_k$ up to $n$ and $\| \cdot \|$ is Euclidean norm, $$ \|a_{n+1} - a_n\| = o(1/n^{\epsilon}) \Rightarrow \|\bar{a}_{n+1} - \bar{a}_n\| = o(1/n^{1+\epsilon})? $$
It should be intuitively, but I can't prove it. This (if true) will imply that $\bar{a}_n$ will converge to a point, right?
The entire thing is false if $\epsilon<1$. Let's say WLOG they're real numbers. Put $r\in(\epsilon,1)$, and $a_n=n^{1-r}$, so your condition is clearly satisfied since $a_{n+1}-a_n=n^{1-r}[(1+\frac{1}{n})^{1-r}-1]=(1-r)n^{-r}+O(n^{-r-1})$. Then $n\overline{a}_n=\sum_{k=1}^nk^{1-r}$ satisfies (similar argument to the proof of the integral test) $$ \frac{(n+1)^{2-r}-1}{2-r}=\int_1^{n+1}x^{1-r}dx\geq \sum_{k=1}^n k^{1-r} \geq \int_0^{n}x^{1-r}dx=\frac{n^{2-r}}{2-r}.$$
Hence $\overline{a}_n = \frac{1}{2-r}n^{1-r}+O(n^{-r})$ so we obtain once more $\overline{a}_{n+1}-\overline{a}_n=\frac{1-r}{2-r}n^{-r}+O(n^{-r-1})$, which is not $o(n^{-1-\epsilon})$. In particular, $\overline{a}_n$ is divergent.
A similar result can be proven for $\epsilon>1$, but in that case, you do get convergence for $a_n$ (since the sequence is Cauchy) and then for $\overline a_n$. (The case $\epsilon=1$ can yield different things.)
PS: Please write your questions more clearly.