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So I need to find the infimum of$\{\frac{1}{n^2} | n \in \Bbb N\}$

I know that this means that I need to find some $x$ where $x < \frac{1}{n^2} \forall n \in \Bbb N$.

By intuition I know that $\lim_{n \to \infty}$, is $0$, which means that the infimum should be $0$, but is there a more formal way of proving that without using limits?

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    Clearly $\frac{1}{n^2}> 0$ for all $n > 0$. For $a > 0$ you can find $n > 0$ such that $a > \frac{1}{n^2}$ and $a$ isn't a lower bound for the set. Now by definition $0$ is the infimum (largest lower bound) of the set. – 2017-02-22
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    @MatiasHeikkilä I reached the same conclusion, and the part that I wasn't so sure about was saying that "you can find an $n > 0$ such that $a > \frac{1}{n^2}$" Do I need to somehow prove that statement? I know it's true intuitively, but is there some formal property of natural numbers or anything else I can base that on? Do I even need to be that rigorous? – 2017-02-22
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    Well you can solve the inequality: $a > \frac{1}{n^2}$ if and only if $n > \frac{1}{\sqrt{a}}$. Any $n$ satisfying this condition suffices (we do believe there are arbitrarily large natural numbers, right? ;) ) – 2017-02-22
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    @AR7 That comes via archimedean property. – 2017-02-22

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Maybe this can help. Let us write $$A=\bigg\{\frac{1}{n^2}:n\in\Bbb N\bigg\}.$$ Clearly, $0$ is a lower bound of $A$. So, the set $A$ is a non-empty set bounded below. Thus, we can find a real number $w$ such that $$\inf A=w.$$ Then, $$w\geq 0.$$ Let $\epsilon >0$. Then, using the Archimedean Property, we can find $n\in\Bbb N$ such that $$\frac{1}{n}<\sqrt{\epsilon}.$$ Since $w$ is a lower bound of $A$ and $\frac{1}{n^2}\in A$, we get $$w\leq\frac{1}{n^2}<\epsilon.$$ Hence, $$w< \epsilon\quad \forall \epsilon>0.$$ Thus, $$w\leq 0$$ and hence $$w=0.$$

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We already know that $0<\frac{1}{n^2}$ for any $n\in\mathbb{N}$. By the definition of infimum you only need to show that $0$ is the biggest number with this property.

Assume it is not true, so there exists $x>0$ such that $x < \frac{1}{n^2}$ for any $n\in\mathbb{N}$. Then $xn^2<1$ for any $n\in\mathbb{N}$.

Now without a loss of generality we may assume that $x<1$. Otherwise $n=1$ would contradict the inequality. Put $n := \lceil\frac{1}{x}\rceil$ (here $\lceil\cdot\rceil$ denotes the ceiling function). Note that $n\in\mathbb{N}$ and furthermore

$$xn^2=x\bigg\lceil\frac{1}{x}\bigg\rceil^2 \geq x\bigg(\frac{1}{x}\bigg)^2=\frac{1}{x}>1$$

Last inequality since $x<1$. Contradiction. $\Box$.

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    It works, but I would say it's easier to just say that for all $n \geq 1/\sqrt{x}$ the inequality is not satisfied. No need to use floor functions. – 2017-02-22
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    @TMM Right, so the floor function is simply to give explicite construction of $n$, without saying "there exists a natural bigger then $\frac{1}{\sqrt{x}}$". – 2017-02-22
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    "The first integer bigger than $1/\sqrt{x}$" is also an explicit construction of $n$, again avoiding the floor function. – 2017-02-22
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    @TMM But that's exactly what I did. You just have to recall how the floor function is defined. Well, almost, I could use the ceiling function instead, but for nonintegers $\lceil r \rceil=\lfloor r\rfloor + 1$. Yeah, ceiling will be better, more readable. Fixed. – 2017-02-23
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    I know what the floor and ceiling functions are; I'm just saying it is unnecessary to introduce new notation. And I still don't get why you're setting $n = 1/x$ instead of $n^2 = 1/x$ - sure, you could also set $n = 712/x^2$, but that's also not the most natural, tightest choice. – 2017-02-23
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    @TMM I don't care if the choice is natural or not. It was natural to me. What does it even mean? The proof is correct, nothing else matters in maths. BTW by the same argument $n:=1/\sqrt{x}$ is bad because you introduce new notation of $\sqrt{\cdot}$. I don't really see how is that better then introducing the floor/ceiling funciton. You really are nitpicking. – 2017-02-23
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Hint:

prove it indirectly, assume there is a bigger lower bound: $\epsilon >0$.

Now what can you say about $\frac{1}{n^2}$ where $n=\lfloor\sqrt{1/\epsilon}\rfloor+1$?

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There is no inf that is in the set, because all values are positive but for any value, there is a smaller one.

However, there is a lim inf, and it is zero, because, for any $c> 0$, all values beyond a $n$ that depends on $c$ are within $c$ of zero.

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    Infimum of a set can be outside of the set. – 2017-02-22
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    There [is](https://en.wikipedia.org/wiki/Least-upper-bound_property) an inf since the set is bounded from below. – 2017-02-22
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    I edited it to reflect what I meant. – 2017-02-22
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    Inf has a very widely accepted definition you aren't using. – 2017-02-22