# Order property of infinite series

I had a question about a result that I found myself trying to use often.

Suppose $$\{a_n\}$$ and $$\{b_n\}$$ are sequences in $$[0,\infty]$$ such that $$a_n \leq b_n$$ for all $$n \in \mathbb{N}$$. Then,

$$$$\sum_{n = 1}^\infty a_n \leq \sum_{n = 1}^\infty b_n$$$$

I know that this result holds for sequences taking values in $$[0,\infty)$$, so my question is really about if allowing $$\infty$$ to appear in the sequence changes the results. I think that the same proof holds here, namely

$$$$\sum_{n = 1}^\infty a_n = \lim_{N \to \infty} \sum_{n = 1}^N a_n \leq\lim_{N \to \infty} \sum_{n = 1}^N b_n = \sum_{n = 1}^\infty b_n$$$$

For clarity, I will also note that I'm using the convention that for any $$x \in [0,\infty)$$, $$x < \infty$$ and $$x + \infty = \infty$$.

• Yup, the proof would likely be identical. Intuitively, note that, for all except some specified $i$, $a_i$ and $b_i$ are finite. When some $b_i$ is infinite, the sum $\sum b_i$ from then onwards would be infinite, and thus $\sum a_i \le \sum b_i = \infty$ regardless of whatever values $a_i$ takes on. If $a_i = \infty$ for any particular $i$, then $a_i = b_i = \infty$ on the premise of $a_i \le b_i \; \forall i$, and thus $\infty = \sum a_i \le \sum b_i = \infty$. So the inclusion of $\infty$ doesn't seem to pose any problems. – Eevee Trainer Mar 25 at 21:29
• Awesome, I appreciate the confirmation. Your explanation also helps explain it intuitively very well. @EeveeTrainer – Colin Jackson Mar 25 at 21:38