Find and prove the supremum of the following set: $S=\{\frac{n-1}{n^2+5}:n>1, n\in\Bbb N\}$ We are given the following set: 
$$S=\left\{\frac{n-1}{n^2+5}:n>1, n\in\Bbb N\right\}$$
By using the monotonicity argument, I have found that the sequence is increasing when $n \le3$ and decreasing when $n \ge4$. Therefore, its largest value is $x_n = 1/7$. Is my argument correct and how can I prove this assumption? 
I am trying to show that for any fixed $\varepsilon$>0, we should have that $\frac{n-1}{n^2+5} > 1/7 -\varepsilon$, but I am currently stuck at this step. 
 A: Let $f(n)=\frac{n-1}{n^2+5}.$ 
You proved that $f(3)\geq f(n)$ for all $n\leq 3$ and  $f(4)\geq f(n)$ for all $n\geq 4$, which says that $$\max\{f(4),f(3)\}=f(3)=\frac{1}{7}$$ is a maximal value.
A: The answer "The sequence is increasing when $n \leq 3$ and decreasing when $n \geq 4$" is not by itself enough: it doesn't tell you what happens between $n=3$ and $n=4$. However, assuming you've done the calculations, you should have found that actually the value is the same ($\frac{1}{7}$) at both those points. So you're right in your conclusion.
The above has shown that $\frac{1}{7}$ is an upper bound for the sequence; strictly speaking, you need to show that it is the least upper bound. However, that's clear because the sequence actually attains that value. No upper bound for a sequence can be lower than some specific member of the sequence.

For examples of why you need to have done that little bit of extra work: the sequence $1,2,3,2,2,2,\dots$ has supremum $3$, which it attains at $n=3$. The sequence $1,2,3,6,5,4,3,2,2,2,\dots$ has supremum $6$, which it attains at $n=4$. Both these sequences satisfy the requirement that the sequence be increasing on $n \leq 3$ and decreasing on $n \geq 4$.
