Infimum of $\left\{\frac{n}{n^2+1}\:\:;\:n\:\in \mathbb N\right\}$ with a proof Consider $A=\left\{\frac{n}{n^2+1}\:\:;\:n\:\in \:N\right\}$.  I need to find and prove $\inf(A)$.
So I know that I need to prove that for every $\epsilon > 0$ exists some $a\:\in A$ such that $a\:=\:\frac{n_0}{n_0^2\:+1}$ for some $n_0$, so  $\inf(A)\:+\:\epsilon \:>\:a$  
What is the method? How to choose what $\epsilon$ is bigger than?

Revised to match corrected question and the OP’s background. 
Let $a_n=\dfrac{n}{n^2+1}$. Then
$$\begin{align*}
\frac{n+1}{(n+1)^2+1}-\frac{n}{n^2+1}&=\frac{n+1}{n^2+2n+2}-\frac{n}{n^2+1}\\\\
&=\frac{(n+1)(n^2+1)-n(n^2+2n+2)}{(n^2+1)(n^2+2n+2)}\\\\
&=\frac{n^3+n^2+n+1-(n^3+2n^2+2n)}{(n^2+1)(n^2+2n+2)}\\\\
&=\frac{-n^2-n+1}{(n^2+1)(n^2+2n+2)}\;.
\end{align*}$$
The denominator is always positive, so this is negative when the numerator is negative, i.e., when $-n^2-n+1<0$. This occurs when $n^2+n>1$, i.e., when $n(n+1)>1$, and this is clearly true for every positive integer $n$. Thus, we know that $a_1>a_2>a_3>\ldots\;$.
How small can these fractions get? For every positive integer $n$ we have 
$$0<a_n=\frac{n}{n^2+1}<\frac{n}{n^2}=\frac1n\;.\tag{1}$$
Thus, $a_n$ is always bigger than $0$, but by taking $n$ big enough, we can get it as close to $0$ as we want. That says that $\inf A$ ought to be $0$. Now you just have to show that if $\epsilon>0$, there is an $n$ such that $a_n<\epsilon$. The inequality $(1)$ should help you do that.
