# 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.

• Just write it as $\frac{n-1}{n^2+5}=\frac{1}{7}-\frac{(n-3)(n-4)}{7n^2+35}$. This shows both that it is always (for $n\in\mathbb{N}$) less than $1/7$, since the term being subtracted is always $\geq0$, and that you get $1/7$ for $n=3$ and $n=4$. Commented Feb 1, 2018 at 17:22

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.

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 $n \geq 4$.

• Thanks! I got the idea. But how can I rigorously prove this by using the epsilon definition? Commented Feb 1, 2018 at 17:05
• Can I clarify: you're satisfied that every value of the sequence is less than or equal to this purported supremum, and now you wish to show that it is indeed a supremum in the sense that there are no smaller upper bounds? Commented Feb 1, 2018 at 17:15
• Yes, that's where I am stuck at the moment. Commented Feb 1, 2018 at 17:18
• I've edited something in to address that; to use $\epsilon$ is definitely overkill here. You can show directly that anything smaller than $\frac{1}{7}$ fails to be an upper bound for the sequence, because (for instance) the third member of the sequence is bigger than the "upper bound" you're trying. Commented Feb 1, 2018 at 17:20