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?
On
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.
On
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. $\Box$.
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 $$w=0.$$