Basic problem with supremum question.

[tamintl]

Hey guys!

Revising for an exam and I've come across a pretty basic problem.

Question: Prove that the supremum of the set A : { 3n / (5n+1) :n€N} is 3/5

My answer: So 3n / (5n+1) ≤ 3n / 5n = 3/5 so 3/5 is an upper bound.

Now, We claim that 3/5 is the least upper bound. Take β < 3/5 so now I need a positive integer n > .........This is bit I don't know how to do.... (how do I choose this part?)
I then know that you re-arrange n>........ to see that the β we chose earlier is: β < 3n / (5n+1) which is impossible, hence Sup(A) = 3/5

I hope you understand what I mean..
Regards as always
Tam

[radou]

You should rearrange your inequality somehow to show that for any given ε > 0, you can find some natural number N such that 3N / (5N + 1) > 3/5 - ε. I haven't tried it out myself, but this should be the concept, unless I'm mistaken.

[tamintl]

You should rearrange your inequality somehow to show that for any given ε > 0, you can find some natural number N such that 3N / (5N + 1) > 3/5 - ε. I haven't tried it out myself, but this should be the concept, unless I'm mistaken.

Are you sure? The solutions I have are vague to say the least, but would you have any idea how I could do this?

[HallsofIvy]

Solve the inequality
\frac{3N}{5N+1}> \beta
for \beta< 3/5
or, equivalently,
\frac{3N}{5N+1}> \frac{3}{5}- \epsilon
for N.