Could anybody please clear the subject of proving limits with the epsilon-N technique for sequences?

I understand the definition of a limit,

if a=lim <SUB>n->oo</SUB> a_n

then for every epsilon>0 exists N such as for every n>N

|a-a_n| < epsilon

right?

So now how do you prove that, say, lim <SUB>n->oo</SUB> a_n=0
where a_n=1/n²

I know how to prove it, but I don't understand how exactly it proves it.

what does N>sqrt(1/epsilon) give?

note: _oo=infinity

um...

I really need this. Please, somebody! :(

It's basic calculus!

Right, looks like I'm gonna have to be the one.

let n = N + k; k > 0

Then a_N-a_n = 1/N² - 1/(N + k)²

= (N² + 2Nk + k² - N²)/(N²(N + k)²)

= 2k(N + k)/(N²(N + k)²)

= 2/(N²(N/k + 1))

If the limit exists, then there exists epsilon such that:

|2/(N²(N/k + 1))| < epsilon

No problem; define epsilon = 2/N²

Solving for N:

N = (2/epsilon)^1/2

Now, clearly no matter what (nonzero) value of epsilon you pick, there will be a corresponding value of N. Therefore, a limit exists in this case.




As to how you show the limit is actually 0...

Assume it's a nonzero positive number, m. Then there is an N such that 1/N² = m. Which implies that for N+1 the result of 1/N² is less than m. Which means m is not a limit. Similar for nonzero negative numbers. Ergo, limit is 0.

first,

THANK YOU FOR ANSWERING!!!!!

I OWE YOU!!!!

what do you want me to do? Rob a bank for you? ANYTHING!!!

phew...

Ok, let me get it... the fact that N can be expressed with any epsilon>0 proves that there is a limit to the given sequence?

I'll accept a modest sacrifice. A lamb, maybe? :D

Originally posted by BloodSuckingGerbile
Ok, let me get it... the fact that N can be expressed with any epsilon>0 proves that there is a limit to the given sequence?

Sure, just look at the definition you provided. All I did was follow the definition and show that "for every epsilon>0 there exists N such as..."

Well, perhaps your wording was a little misleading. It should start: "if, and only if a=lim...". IOW the epsilon/N condition is both necessary and sufficient. So if you show the condition holds then the limit exists.

Intuitively the definition makes a lot of sense. After all, if a series tends to a limit then after a while all of its remaining terms should lie within a certain distance (epsilon) from that limit. It can oscillate and do wild things, but eventually it has to settle down around the limit. Which is basically all the definition is trying to say. As epsilon gets arbitrarily small the series approaches the limit assymptotically.

I'll find the best lamb I can, young fat and juicy. You're going to love it :D

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Ok. I think I got it.

But... how do you prove that, say 2^{(-1)^n} or sin(n) has no limit with the epsilon - N technique? By assuming it exists?

Oh, and one more question.

The limit of the sequence a_n=n²+1 is infinity, right? If yes, how do you prove it? I mean, you can't write

|a_n-infinity| < epsilon

can you?

Thanks again.

Originally posted by BloodSuckingGerbile
I'll find the best lamb I can, young fat and juicy. You're going to love it :D

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.
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Ok. I think I got it.

But... how do you prove that, say 2^{(-1)^n} or sin(n) has no limit with the epsilon - N technique? By assuming it exists?

Oh, and one more question.

The limit of the sequence a_n=n²+1 is infinity, right? If yes, how do you prove it? I mean, you can't write

|a_n-infinity| < epsilon

can you?

Thanks again.

not A(epsilon>0 : E(N: N>0 : A(n: n>N : |a_n-a|<epsilon)))

= {negation of A and E (De Morgan) }

E(epsilon>0 : A(N: N>0 : E(n: n>N : |a_n-a|>=epsilon)))

So, you have to find a certain epsilon (for instance 1/10) such that for any given N you can find a n (as a function of N) such that
|a_n-a|>=epsilon.

Finding this epsilon value and function to calculate n is now your task. It is entirely dependent of the shape of a_n.

For a_n=sin(n) epsilon = 0.7 and n(N) = round(2*N*pi + pi/2) are good choices I think.


For the infinity-case you actually have a different definition of the limit because infinity is not a number. What we mean is that the sequence can become arbitrarily large so the definition would be:

A( delta: delta >0 : E(N: N>0 : A(n: n>N : a_n >delta )))


Greetings,

Han.

ok.

Thanx, dude.
I think I got it thanks to you and Overdoze.

I owe you both now.

...

aam... one more question:

how do you post the quote with the "originally posted by [username]" title? I just don't want to start a thread for that...

:)

use the <button>quote</button> button just above the <button>Post reply</button> button.

you mean "submit reply"?

There are no buttons above "Submit Reply".