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 $x < \frac{1}{n^2} \forall n \in \Bbb N$.
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?
 A: 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 $$w=0.$$
A: 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. $\Box$.
A: 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.
A: Hint: 
prove it indirectly, assume there is a bigger lower bound: $\epsilon >0$.
Now what can you say about  $\frac{1}{n^2}$ where $n=\lfloor\sqrt{1/\epsilon}\rfloor+1$?
