# SUP question

1. Nov 25, 2008

### transgalactic

S={3/2,5/3,7/4,9/5,11/6 ...}
the formula for this series is S(n)=2n+1/n+1
the limit for it as n->infinity gives me 2

i show that 2 is upper bound

2n+1/n+1<2 => 2>1 (always true)

now i need to show that 2 is the "least upper bound".
if 2 is not the least upper bound then there is a certain "x" for which
2n+1/n+1<2-x (2-x is the least upper bound)

now they are doing some thing really odd
2-x<2n+1/n+1

why???
this move is illegal
2-x cannot be smaller
its supposed to be the "least upper bound"

??

2. Nov 25, 2008

### transgalactic

ok i found out why they do this move

2-x<2n+1/n+1
its to prove that 2-x is not the least upper bound.
they develop this innequality to this point
n>1/x -1

and here they conclude that 2-x is not the least upper bound.
why they conclude that 2-x is not least upper bound?

3. Nov 25, 2008

### Staff: Mentor

You're obviously in a fairly high mathematics class, so why would you write 2n + 1/n + 1, when you almost certainly mean (2n + 1)/(n + 1).

It would be legitimate to interpret what you've written as:
$$2n + \frac{1}{n} + 1$$. If you write rational expressions on one line, put parentheses around the terms in the numerator and around those in the denominator.

You're not a new member of this forum, so it might behoove you to learn some LaTeX to format what you post something like this:
$$S(n) = \frac{2n + 1}{n + 1}$$
Your conclusion above is certainly true, but it's not at all clear that (2n + 1)/(n + 1) < 2 implies this conclusion. You need to show that the first inequality is true.

4. Nov 25, 2008

### Staff: Mentor

They are doing a proof by contradiction. One way to prove that p ==> q, (where p and q are statements) is to assume that p is true and q is false. If you arrive at a contradiction, that means your assumption that q was false must actually have been false. IOW, q must be true.

5. Nov 25, 2008

### transgalactic

i know that its a proof by contradiction
i cant understand the last step of it:

i got
n>1/(x -1)
they say
"since n exists satisfying the above inequality our claim is proves
2 is Least Upper Bound"

i cant see the logic of this line
can you explain it in simpler words?

6. Nov 25, 2008

### transgalactic

there may be a "n" that satisfy our contradiction
but there maybe another "n" whose not

??

7. Nov 25, 2008

### HallsofIvy

Staff Emeritus
There exist positive x so that is true. And you want to prove that statement is NOT true.

Since you haven't show exactly what "they" are saying, I can only guess that this is what they want to prove.

IF there exist such an x it can't. But they want to prove that there is NO such x.

What is "supposed to be the "least upper bound" "? Not x certainly! It is "2" that they are trying to prove is the least upper bound.

(2n+1)/(n+1)= 2- 1/n. Given any positive number x, 1/x is a positive real number so, by the Archimedian property, there exist an integer N such N> 1/x. Then 1/N< x, -1/N> -1/x and 2- 1/N> 2- 1/x. Thus, 2- 1/x cannot be an upper bound on (2n+1)/(n+1) for any positive x.

8. Nov 25, 2008

### transgalactic

i know that the i thought as "odd" was for prooving that x-2 is not LUB
if this contradiction expression is proved then 2-x is not a LUB(then 2 is LUB)
we came to the last part

n>1/(x -1)

i know that n can be :0 1 2 3 5 ... infinity
i dont know what the properties of X(in the article its epsilon)

so without that last piece about "what numbers could be taken by x"
i cant say that this expression is true

n>1/(x -1)
??

Last edited: Nov 26, 2008
9. Nov 25, 2008

### Vid

What are you even asking? Both the solution Halls gave and the solution in that link explain everything rather clearly.

10. Nov 26, 2008

### transgalactic

i cant understand the lst part of the proof

why does this expression is valid?
n>1/(x -1)

11. Nov 26, 2008

### transgalactic

i cannot understand this words of hallsofivy
"(2n+1)/(n+1)= 2- 1/n. Given any positive number x, 1/x is a positive Real Number so, by the Archimedian property, there exist an integer N such N> 1/x. Then 1/N< x, -1/N> -1/x and 2- 1/N> 2- 1/x. Thus, 2- 1/x cannot be an upper bound on (2n+1)/(n+1) for any positive x.
"

is there simpler expanation?

12. Nov 26, 2008

### transgalactic

and in the article "Since there exist n ∈ N satisfying the above inequality.."

why its satisfying the inequality ??

13. Nov 26, 2008

### transgalactic

how it shows??

http://img354.imageshack.us/img354/6885/40420484iw4.gif [Broken]

Last edited by a moderator: May 3, 2017
14. Nov 26, 2008

### Office_Shredder

Staff Emeritus
By basic inequality arithmetic?

$$n > \frac{1}{ \epsilon } -1$$ add 1 to both sides

$$n+1 > \frac{1}{ \epsilon }$$ raise both sides to the (-1) power (which flips the inequality)

$$\frac{1}{n+1} < \epsilon$$ add 1 to both sides and subtract epsilon, 1/(n+1) from both sides

$$1 - \epsilon < 1 - \frac{1}{n+1}$$

then notice

$$1-\frac{1}{n+1} = \frac{n+1}{n+1} - \frac{1}{n+1} = \frac{n}{n+1}$$

This is pretty easy stuff, and it sounds like you need to review your basic inequality arithmetic

15. Nov 26, 2008

### transgalactic

ok what next

1-e<n/(n+1)

so what makes this expression true?

16. Nov 26, 2008

### HallsofIvy

Staff Emeritus
$$1 - \epsilon < 1 - \frac{1}{n+1}$$
and then
$$1-\frac{1}{n+1}= \frac{n}{n+1}$$
Therefore
$$1-\epsilon< \frac{n}{n+1}$$

17. Nov 26, 2008

### transgalactic

my proffesor said that there is no general proof ,he said that
when i develop this expression
n>1/(e -1)
i say
"for every epsilon that could be inputed
i can put an "n" which sutisfies this expression"

is that correct?

18. Nov 26, 2008

### transgalactic

i think i got it

if n>1/(e -1)

then 1-e is not the least upper bound
and if there is one case where its not the least upper bound ("n>1/(e -1)") then its not
the least upper bound at all
am i correct?