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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,

\begin{equation} \sum_{n = 1}^\infty a_n \leq \sum_{n = 1}^\infty b_n \end{equation}

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

\begin{equation} \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 \end{equation}

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

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    $\begingroup$ 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. $\endgroup$ – Eevee Trainer Mar 25 at 21:29
  • $\begingroup$ Awesome, I appreciate the confirmation. Your explanation also helps explain it intuitively very well. @EeveeTrainer $\endgroup$ – Colin Jackson Mar 25 at 21:38

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