Proving 2 as the Least Upper Bound for the Series S(n) = 2n+1/n+1

[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"

??

 

[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?

 

[Mark44]

S={3/2,5/3,7/4,9/5,11/6 ...}
the formula for this series is S(n)=2n+1/n+1

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}

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)

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.

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"

??

 

[Mark44]

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?

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.

 

[transgalactic]

i know that its a proof by contradiction
i can't 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 can't see the logic of this line
can you explain it in simpler words?

 

[transgalactic]

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

??

 

[HallsofIvy]

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)

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

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

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

why?
this move is illegal
2-x cannot be smaller

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

its supposed to be the "least upper bound"

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.

 

[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 don't know what the properties of X(in the article its epsilon)
https://www.math.purdue.edu/academic/files/courses/2007fall/MA301/MA301Ch6.pdf (on page 4)

so without that last piece about "what numbers could be taken by x"
i can't say that this expression is true

n>1/(x -1)
??

 

[Vid]

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

 

[transgalactic]

i can't understand the lst part of the proof

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

 

[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?

 

[transgalactic]

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

why its satisfying the inequality ??

 

[transgalactic]

how it shows??

http://img354.imageshack.us/img354/6885/40420484iw4.gif

 

[Office_Shredder]

By basic inequality arithmetic?

n &gt; \frac{1}{ \epsilon } -1 add 1 to both sides

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

\frac{1}{n+1} &lt; \epsilon add 1 to both sides and subtract epsilon, 1/(n+1) from both sides

1 - \epsilon &lt; 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

 

[transgalactic]

ok what next

1-e<n/(n+1)

so what makes this expression true?

 

[HallsofIvy]

You already have that
1 - \epsilon &lt; 1 - \frac{1}{n+1}
and then
1-\frac{1}{n+1}= \frac{n}{n+1}
Therefore
1-\epsilon&lt; \frac{n}{n+1}

 

[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?

 

[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?