Find and justify the Supremum of the following set

Question

Find the supremum of the following set:

$$A:=\left\{{(n-1)\over(2n+3)} : n \in\mathbb{N}\right\}$$

So I have as my answer that sup(A) = ½ but need to justify. We were taught to justify in two steps, first show our answer is an upper bound for the set then show it is a least upper bound.

For step 1: if a is a member of A then a = (n-1)/(2n+3) for some natural number n and (n-1)/(2n+3)

Step 2 is where I am uncertain. I think I need to show that for any y<½ there exists a member of the set between y and ½ but not sure how to do this. Thanks for all the answers so far!

@Fkins The second answer in the closed (hold) first post looks pretty clear and neat to me. Anything you didn't understand there ? — DonAntonio, Oct 27 '16 at 20:42
@DonAntonio: The post is not deleted, only closed (resp. on hold). — celtschk, Oct 27 '16 at 20:45
@DonAntonio I haven't learnt monotone sequences or finding limits yet so did not understand the answer. Any advice on proceeding from showing it is an upper bound to a least upper bound with out using above knowledge? Thanks! — Fkins, Oct 27 '16 at 20:47
@Fkins It's hard to understand how you're given this question, on sequences, without knowing what is monotone sequences... The only way I can think of is what fleablood wrote in the above comment...perhaps. — DonAntonio, Oct 27 '16 at 20:48
1/2 - (n-1)/(2n+3)= (2n+3-2n+2)/2 (2n+3)=5/2 (2n+3). Can you argue using that? — fleablood, Oct 27 '16 at 20:48
You can calculate the derivate, verify where it is positive and additionally calculating the limit for $x\rightarrow +\infty$. — Peter, Oct 27 '16 at 20:57
@Peter The OP said he hasn't yet studied monotone sequences or even limits. It seems like a long shot to assume he knows derivatives... — DonAntonio, Oct 27 '16 at 20:59

score 0 · Answer 1 · answered Oct 27 '16 at 21:04

0

You want to show $$\frac{n-1}{2n+3}\le \frac{1}{2}$$ for all $\ n\in \mathbb N$

The above inequality is equivalent to $$2n-2\le 2n+3$$ if $\ 2n+3\ $ is positive (in particular if $n$ is non-negative).

Since the second inequality is obviously true , you have shown that no member of the sequence can exceed $\frac{1}{2}$. You only need the limit to show that $\frac{1}{2}$ is even the supremum.

answered Oct 27 '16 at 21:04

Peter

84,454

If you do not know the value of the (possible) supremum or infimum (usually the case) , begin with the limit and try to show that it is a supremum or infimum. – Peter Oct 27 '16 at 21:10

score 0 · Accepted Answer · answered Oct 27 '16 at 21:12

Let $a\in A $.

Then $a=\frac {n-1}{2n+3} $ for some $n $

$a =\frac {n-1}{2n+3}<\frac {n-1}{2n-2} =\frac 12$

So $\frac 12$ is upper bound of $A $.

Step 2:

Suppose $x < \frac 12 $. Can we prove that there is an $n $ so that

$x < \frac {n-1}{2n+3} < \frac 12$?

$x < \frac {n-1}{2n+3} < \frac 12\iff$

$2x <\frac {2n-2}{2n+3} <1\iff $

$2x (2n+3)<2n-2 <2n+3\iff $

$4xn+6x <2n-2\iff $

$2xn+3x <n-1 \iff $

$3x+1 <n (1-2x)\iff $

$\frac {3x+1}{1-2x} < n $ (remember $x < \frac 12 $ so $1-2x>0$)

As $\mathbb N $ is unbounded we can always find such an $n $. So $x$ is not an upper bound.

That makes total sense, thanks for dumbing it down to my level! — Fkins, Oct 27 '16 at 22:56

Find and justify the Supremum of the following set

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