Finding limits without epsilon and delta

[kingstrick]

Homework Statement

I need to prove that 1/n has a limit of zero using the following definition:

The statement that the point sequence p1, p2, . . . converges to the point x means that if S is an open interval containing x then there is a positive integer N such that if n is a positive integer and n ≥ N then pn ∈ S.

and

The statement that the sequence p1, p2, p3, . . . converges means that there is a point x such that p1, p2, p3, . . . converges to x.

Homework Equations

the instructor was adamant that we are not to use epsilon delta in these proofs for our text does not use such notions. But I am at a lost as to how we prove it without them. Do I just substitute a different variable in the epsilon and delta places? Any advice on how to get started would be appreciated.

The Attempt at a Solution

This is my second attempt in a real analysis class :( It is amazing how different the material is from course to course even for the same course.

 

[jedishrfu]

Perhaps this will help:

http://www.mathcs.org/analysis/reals/numseq/

They mention that you can prove convergence if you can show its bounded and monotonic.

 

[kingstrick]

I know but we haven't 'learned' bounding or monotonic yet. All we have is this definition to go by.

 

[jedishrfu]

I know but we haven't 'learned' bounding or monotonic yet. All we have is this definition to go by.

Then you must find something in the definition that will help you. You could also try rereading your book to see if there is something in the discussion or even in the problem sets that might give you a clue.

 

[HallsofIvy]

To prove that 1/n converges to 0, show that if U is some open interval containing 0 then there exist N such that if n> N then 1/n is in U.

Since U is an open interval containing 0, there exist some number, d, such that the interval (-d, d) is a subset of U. Now show that, for sufficiently large N, if n> N then 1/n is in (-d, d). (Since n> 0, that is equivlent to 1/n< d. How large must n be?)

 

[kingstrick]

n>d ?

 

[Ray Vickson]

n>d ?

Well, suppose d = 0.0001. How large must n be to have 1/n < 0.0001? You claim n > 0.0001 will do. Do you really think that?

 

[kingstrick]

Well, suppose d = 0.0001. How large must n be to have 1/n < 0.0001? You claim n > 0.0001 will do. Do you really think that?

I see what you mean.

in your case n would have to be >10000

so then... n > 1/d

 

[kingstrick]

To prove that 1/n converges to 0, show that if U is some open interval containing 0 then there exist N such that if n> N then 1/n is in U.

Since U is an open interval containing 0, there exist some number, d, such that the interval (-d, d) is a subset of U. Now show that, for sufficiently large N, if n> N then 1/n is in (-d, d). (Since n> 0, that is equivlent to 1/n< d. How large must n be?)

What always stumbles me here though, is what exactly is N? Is it like the Nth term in my sequence? Or is it any random value N, there is a nth term (referring to little n) to the right?

 

[kingstrick]

So here is my solution to a similar problem, is my solution/approach correct?

Prove that M = {1, 1+1/2, 1+1/4, 1 + 1/6, ...} has a limit point of 1.

Let M = {1, 1+1/2, 1+1/4, 1 + 1/6, ...} and U be some open interval containing 1. Since U is an open interval (a,b) where a<1<b, there exists some d \neq1 in (1,b) such that a \leq (a+d)/2 <1 < (b+d)/2 \leq b. Observe that M is \geq 1and ((a+d)/2, (b+d)/2) is an open interval and a subset of U. Let n be a natural number and n>N where N \in R and 1+1/n < d. Then 1/n < d-1 < d. Then n> 1/d.

Therefore 1 is a limit point.