# Confusion around the way infinite series are defined, and how they are used

According to Wikipedia (https://en.wikipedia.org/wiki/Series_(mathematics)), what we call the sum of an infinite series is really just a shorthand for its limit. E.g. the sum of $$1/2+1/4+1/8+ ...$$ is 1. This is the number that the series tends to, the smallest number that the partial sums will never reach, no matter how many terms you add together. Conceptually, this seems much easier to grasp than actually "adding all the terms up". However, this seems to be in opposition with other ideas in maths. For example, 0.999... is said to be a complete number (one of the main misconceptions about it is that it is growing or expanding). But doesn't this imply that you can "add all the terms up", contrary to the careful way the sum of an infinite series is defined? By "terms" I mean this (I have often seen this definition of 0.999.. in the proofs that it is equal to 1): $$\frac {9}{10}+\frac {9}{100}+\frac {9}{1000}+\frac {9}{1000}+...$$However, I am wary of defining it in this way because it seems like there is no way to truly assigning a value to this series — the best you can do is say that it's limit is 0.999..., which seems different to saying that it is 0.999... This is similar to how I don't quite follow the leap made in the resolution of Achilles and the Tortoise, where the limit of $$1/2+1/4+1/8+1/16+...$$ is 1, so consequently the tortoise can be caught up. I feel as if the concept of making "infinite steps" is strange, especially when it seems like the definition of an infinite series has circumvented this problem with the idea of limits. I'm sorry if this question sounds like it has turned into many; I'm just genuinely curious about infinity and the role it plays in mathematics. Thanks

• $0.9999...$ means that series. Aug 15, 2019 at 16:33
• Regarding $1/2+1/4+1/8+\cdots$, consider this: You walk towards a door. First step is half the distance, next step is half the remaining distance, etc. Clearly, you never pass the door. Moreover, there's no point in front of the door you don't eventually pass. If we want to assign a value to "$1/2+1/4+1/8+\cdots$", then that value can't be larger than $1$, nor smaller than $1$. With no other options, we agree to say that the value is exactly $1$. Likewise for $0.999\ldots$: it's not greater than $1$, yet it surpasses everything less than $1$; we agree to say it's exactly $1$.
– Blue
Aug 15, 2019 at 16:52
• @Blue Thanks for answering. When you say "it's exactly 1", do you mean that the sum tends to 1? I still feel uncomfortable saying that it 1, because again, I can't see a way of you getting to the door, let alone past it.
– Joe
Aug 15, 2019 at 17:00
• @Joe that's what limits allow you to do. Without ever uttering "limit," you'd be right. Aug 15, 2019 at 17:15
• In the words of Sherlock Holmes: "When you have eliminated the impossible, whatever remains, however improbable, must be the truth." In the door scenario, we have ruled-out passing the door or stopping before it; these are impossible. The sole remaining option, however weird it might seem, is that infinitely-many steps land us at the door. This is more-or-less what series (and, more generally, limits) do: rule-out every value except one, and we take that one value to be what the series (or limit) represents.
– Blue
Aug 15, 2019 at 17:22

We define $$0.999\dots$$ as the limit of this geometric series, which of course is $$1$$. This trips up a lot of people, but it shouldn't. It makes just as much sense to write$$0.999\dots=\sum_{k\ge1}\frac{9}{10^k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{9}{10^k}=1$$in base 10 as to write$$0.111\dots=\sum_{k\ge1}\frac{1}{2^k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{2^k}=1$$in binary.