How to prove $B$ is the least possible upper bound?

Question

I saw the following sequence-

$S_n=\frac12,\frac23,\frac34,\frac45,\frac56,\ldots,\frac{n-1}{n}$.

Now, I may say that-

$$ (A)\leq S_n\leq(B),$$

where, $A=\frac12$ and $B=1$

Now, am I correct in saying that $B$ is the least possible upper bound?

If yes, how do I prove it?

Thanks for any help.

Answer 1

If I'm parsing your question properly, you are asking why $1$ is the least upper bound of the set of values of $\frac{n-1}{n}$.

Take any number $x < 1$, and note that we can find an $n$ such that $\frac{1}{n} < 1 - x$, or, in other words $x < 1 - \frac{1}{n} = \frac{n-1}{n}$. Therefore, $x$ cannot be an upper bound for your set.