My professor gets upset when we use Calculus to solve problems in his Introduction to Abstract Math course. I have taken Linear Algebra, Discrete Math, and Calculus I.
I have no idea how to find the supremum of the sequence $(x_n)$ with $x_n = \frac{2n}{6n+3}$ for $n \geq 1$ and $n \in \mathbb{Z}^+$ without using a limit. I know that the limit of $x_n$ as $n$ approaches infinity is $\frac{1}{3}$, which would be the supremum, but I do not know how to show that without doing a limit. How would I go about finding that?
Best Answer
The transformations suggested above , need not be followed by a limit argument.
Indeed, from $a_n = \frac 13 - \frac{1}{6n+3}$, it is clear that $\frac 13$ is an upper bound. To see that no smaller number is an upper bound, simply consider $\frac 13 - \epsilon$ for any $\epsilon > 0$, and we want to show that this is not an upper bound.
For this, we need an $N$ such that $a_n > \frac 13 - \epsilon$, which translates to $\frac 13 - \frac 1{6n+3} > \frac 13 -\epsilon$, which is equivalent to $\frac 1{6n+3} < \epsilon$, which becomes $6n+3 > \frac 1\epsilon$, and finally $n > \frac {1}{6 \epsilon} - \frac 12$.
Now note that if $N = \left\lceil \frac 1{6 \epsilon} - \frac 12\right \rceil + 1$ (we do ceil because we need to make $N$ an integer. Furthermore, if $\frac 1{6 \epsilon} - \frac 12$ was already an integer, then this ceil doesn't give us a strictly bigger quantity than it, so we add $1$) then $N > \frac {1}{6 \epsilon} - \frac 12$ , so $a_N > \frac 13 - \epsilon$, therefore $\frac 13 - \epsilon$ cannot be an upper bound.
We may have mimicked what we do in a limit argument, but nowhere was the word limit used.
The reason for picking $\frac 13$ as a candidate supremum though, certainly required limit-type intuition, because you have to anticipate that the fraction $\frac 1{6n+3}$ gets closer to $0$ as $n$ increases, which means that $a_n$ gets closer to $\frac 13$.
But it is not possible to solve this problem unless you have a candidate for the supremum, and you can't get a candidate for the supremum without studying what happens to $a_n$ for large $n$, which is like studying the limit.
Therefore, you cannot avoid the entire limit-type approach if you want to solve this problem , the above rephrasing is the best you can do.