How to formally prove that a value is a bound of a sequence? Terrible question, sorry:
Consider the sequence in $\mathbb{R}$, $(\frac{10}{n})_{n \in \mathbb{N}}$. Intuitively, it's clear that a lower bound of this sequence is $0$ and an upper bound is $10$. How could I formally prove (say, analytically or number theoretically) that for any value $x$ in the range of the sequence, $0 \leq x \leq 10$?
 A: Let $n\in \Bbb N$ and $x_n = \frac{10}n$, on the one hand
$$x_{n+1}=\frac{10}{n+1}\leq \frac{10}{n} = x_n,$$
i.e. $(x_n)_{n\in \Bbb N}$ is decreasing. It follows that 
$$x_{n}\leq x_{n-1}\leq \ldots \leq x_1  = \frac{10}{1}=10.$$
On the other hand $10>0$ and $n>0$ so that $$x_n=\frac{10}{n}>0.$$
Hence $$0< x_n \leq 10, \qquad \forall n \in\Bbb N$$
i.e. $10$ is an upper bound of $(x_n)_{n\in\Bbb N}$ and $0$ is a lower bound of $(x_n)_{n\in\Bbb N}$.
A: Fix some $n \in \mathbb{N}$. Since $1/n \leq 1$, we have that $$\frac{10}{n} \leq 10$$  On the other side, we have that $n > 0$, so also $1/n > 0$ and since $10 > 0$ we have that $$0 < \frac{10}{n}$$ In conclusion $$0 < \frac{10}{n} \leq 10$$ Hence take the rule:

Fix some $n \in \mathbb{N}$ and try to bound each term $x_n$ separately. Sometimes you will get a bound which holds for all terms.

A: I would go with induction:


*

*For $n=1$ you get $0 < x_1 = \frac{10}{1} = 10 \le 10$.

*Assume the statement holds for $n \ge 1$, and consider 
$$x_{n+1}=x_n \frac{n}{n+1}.$$
Since $0 < \frac{n}{n+1} < 1$, we have
$0 < x_{n+1} < x_n$, meaning that every point in the sequence is positive, and the sequence is monotone decreasing, starting at $10$.

