Show that $(\frac{n}{n^{2}+1} \mid n \in \mathbb{N}^{+})$ is a decreasing sequence

Question

would anyone please explain to me why the inequality 1

Answer 1

I want to clarify the question. What we do in a proof is

1. Stablish an hypothesis (or more than one).

2. Test if the hypothesis, following the rules of mathematics and mathematical logic, is true or false.

Then we have defined the sequence $(a_n)$ such that $a_n=\frac{n}{n^2+1}$. Then our hypothesis to test is: it is the sequence $(a_n)$ decreasing? If $(a_n)$ would be decreasing then it must be true that $a_n\ge a_{n+1}$ for all $n\in\Bbb N$, by the definition of a decreasing function.

Observe that, as we said before, $a_n=\frac{n}{n^2+1}$ and $a_{n+1}=\frac{(n+1)}{(n+1)^2+1}$. Then we write explicitly the inequality $a_n\ge a_{n+1}$ i.e.

$$\frac{n}{n^2+1}\ge \frac{(n+1)}{(n+1)^2+1}$$

Now, following the rules of mathematics we want to see if the above inequality is true. Then we start to do some algebra over the expression to simplify it and see, in a clear way, if the above inequality is true or false.

If the inequality is true then we proved that the sequence $(a_n)$ is decreasing. I hope you see now clear.

Answer 2

"I don't see how the inequality $1<n^2+n$ makes sense of $a_{n+1}< a_n$",

"I still have difficulty in making the connection why $1<n^2+n$ could lead us to $a_{n+1}< a_n$"

More succintly and better suited than the answer you accepted with such enthusiasm: the key to overcome your difficulty was to follow backwards the equivalences of your text. It seems the precise point you were missing was the meaning of these $\iff$ signs. The useful direction of these signs, in that proof, is the reverse one: $\impliedby.$