SUP question: S = {3/2, 5/3, 7/4, 9/5, 11/6, ...}

[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)

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?

 

[pka]

Re: SUP question..

If \(\displaystyle 1 < x < 2\) then find an integer J such that \(\displaystyle J > {\frac {x - 1} {2 - x}}\).
Can you show that \(\displaystyle s_J > x\)?

 

[transgalactic]

Re: SUP question..

i want to go by their proof

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

 

[pka]

Re: SUP question..

Don't you see, that is the same logic?
Because 2 is an upper bound and that shows that if x<2 then x is not an upper bound.
Therefore, 2 is the least upper bound.

 

[transgalactic]

Re: SUP question..

my main problem is in this last part:

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)
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 cant say that this expression is true

n>1/(x -1)
??

 

[PAULK]

Re: SUP question..

Don't you see, that is the same logic?
Because 2 is an upper bound and that shows that if x<2 then x is not an upper bound.
Therefore, 2 is the least upper bound.

If
2n + 1
Sn = ------
n + 1
How about evaluating 2 - Sn?
2n + 1
2 - ------
n + 1

2n + 2 - (2n + 1)
------------------
n + 1

2n + 2 - 2n - 1
----------------
n + 1

1
--------
n + 1

which --> 0, i.e. can be made < any epsilon. So you reason:
Let x be any number less than 2, then x = 2 - e. Then the above calculation shows that some member Sn > x.

 

[Deleted member 4993]

Reformatted the response above for easier reading

Don't you see, that is the same logic?
Because 2 is an upper bound and that shows that if x<2 then x is not an upper bound.
Therefore, 2 is the least upper bound.

If 
     2n + 1
Sn = ------  
      n + 1

How about evaluating  2 - Sn?

    2n + 1
2 - ------
     n + 1

2n + 2 - (2n + 1)
------------------
    n + 1

2n + 2 - 2n - 1
----------------
    n + 1

   1
--------
  n + 1

which --> 0, i.e. can be made < any epsilon. So you reason:
Let x be any number less than 2, then x = 2 - e. Then the above calculation shows that some member Sn > x.