Generalized Sum Preserves Inequality

Theorem
Let $\left({a_i}\right)_{i \in I}, \left({b_i}\right)_{i \in I}$ be $I$-indexed families of positive real numbers.

That is, let $a_i, b_i \in \R_{\ge 0}$ for all $i \in I$.

Suppose that for all $i \in I$, $a_i \le b_i$.

Furthermore, suppose that $\displaystyle \sum \left\{{ b_i: i \in I }\right\}$ converges.

Then $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$.

In particular, $\displaystyle \sum \left\{{ a_i: i \in I }\right\}$ converges.

Proof
First, it is proven that $\displaystyle \sum \left\{{ a_i: i \in I }\right\}$ converges.

Then, the inequality $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$ is well-defined, and hence can be proven.

Proof of Inequality
Suppose that $\displaystyle \sum \left\{{ a_i: i \in I }\right\} > \sum \left\{{ b_i: i \in I }\right\}$.

Then, as the sums converge, there exists a finite $F \subseteq I$ such that:


 * $\displaystyle \sum_{i \in F} a_i > \sum \left\{{ a_i: i \in I }\right\} - \epsilon$

for every $\epsilon > 0$.

So by picking a suitable $\epsilon$, it may be arranged that:

These inequalities together constitute a contradiction, and therefore:


 * $\displaystyle \sum \left\{{ a_i: i \in I }\right\} \le \sum \left\{{ b_i: i \in I }\right\}$