Find infimum of $\{\frac{1}{n^2} | n \in \Bbb N\}$ in $\mathbb{R}$ 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 $x &lt; \frac{1}{n^2} \forall n \in \Bbb N$.

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?
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 $\epsilon &gt;0$. Then, using the Archimedean Property, we can find $n\in\Bbb N$ such that $$\frac{1}{n}&lt;\sqrt{\epsilon}.$$ Since $w$ is a lower bound of $A$ and $\frac{1}{n^2}\in A$, we get 
$$w\leq\frac{1}{n^2}&lt;\epsilon.$$
Hence,
$$w&lt; \epsilon\quad \forall \epsilon&gt;0.$$
Thus, $$w\leq 0$$ and hence $$w=0.$$
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 $c&gt; 0$,
all values beyond a $n$
that depends on $c$
are within $c$ of zero.