I have the following setup, and I'm having trouble showing a (seemingly intuitive) fact about it. Let $E = \{e_n\}_{n\in\mathbb{N}}$ be a countable subset of $\mathbb{R}$, and let $a_n>0$ be a sequence such that $\sum_{n\in\mathbb{N}}a_n$ converges. I'm defining $$ S(x_0; \delta) = \sum_{n,|x_0-e_n|<\delta}a(n),\qquad x_0\in\mathbb{R}\backslash E,\ \delta>0 $$ I'm trying to show that $\lim_{\delta\to 0}S(x_0;\delta) = 0$. It seems trivial (by inspection), but I can't get a solid rigorous argument for why this should be true.
My attempt:
What I attempted is this. Suppose for contradiction that $b = \inf_{\delta>0}S(x_0;\delta) > 0$. Choose $\delta_0>0$ and $N\in\mathbb{N}$ so that $S(x_0;\delta_0) < b+\frac1N$. Then define a monotonically decreasing sequence $\delta_k>0$ so that $S(x_0;\delta_k) < b+\frac1{N+k}$. Then $\delta_k\to 0$, so $$ \bigcap_k(x_0-\delta_k,x_0+\delta_k) = \emptyset $$ implying there are no $e\in E$ that we "can't remove" by making $\delta$ smaller by choosing a larger $k$. In other words, the sum $\lim_{k\to\infty}S(x_0;\delta_k)$ is the empty sum, which is zero, contradicting $b>0$.
I'm not very happy with this proof though, it seems nonrigorous. Any ideas on how to make this more rigorous?
Fix $\varepsilon>0$. Then there exists $n_0$ such that $\sum_{n>n_0}a(n)<\varepsilon$. Let $$\delta=\min\{|x_0-e_n|:\ n=1,\ldots,n_0\}.$$ Then, if $|x_0-e_n|<\delta$, we are guaranteed that $n>n_0$. Thus $$ \sum_{n,|x_0-e_n|<\delta}a(n)\leq\sum_{n>n_0}a(n)<\varepsilon. $$