So let $\lim_{n\rightarrow\infty}a_n = a$ and for simplicity let $\{a_n\}$ be increasing and $a>0$. Now consider $\sum_n (a- a_n)$. I wrote a proof that this doesn't converge, and I'm wondering where I went wrong, or if it just works because we have an increasing sequence (part of the proof relies on it being monotone).
Proof: For any $\epsilon > 0$ we have that there exists an $N$ so that the terms of the series $a-a_n = b_n <\epsilon$ for all $n>N$. In particular consider $\epsilon$ of the form $\frac{1}{k}$ for some integer $k$. We have that for every $k\geq 1$ that there exists an $N$ so that $b_n < \frac{1}{k}$ for all $n>N$.
Now since $\{a_n\}$ is increasing, we know that $b_n$ is bounded above by $a- a_0$. Let $k_0$ be the floor of $\frac{1}{a-a_0}$. Then for every integer $k>k_0$, there exists an integer $M$ so that for all $n\leq M$ we have that $b_n \geq \frac{1}{k}$.
Thus we have $\sum b_n \geq \sum_{k=k_0+1}^\infty \alpha_k\frac{1}{k}$ where the $\alpha_k$ are integer weights. They represent the distance between the $M$ we obtain for each consecutive $k$. In particular, $\alpha_k \geq 1$ for all $k$. Thus we have $\sum b_n \geq \sum_{k=k_0+1}^\infty \frac{1}{k}$ which proves divergence by the comparison test.
This completes the proof. Hopefully this is easy to read!
edit: seems that the critical problem is where I assumed that $\alpha_k \geq 1$ for all $k$. Since $\alpha_k$ = $M_k - M_{k-1}$ (where these $M$ come from paragraph 3) it could easily be the case that $\alpha_k = 0$, if the sequence approaches its limit fast enough.