# Thread: Definition of sequence to infinity

1. ## Definition of sequence to infinity

Hey

Can anyone explain what A and N are in the definition for a limit of infinity in a sequence:

$a_{n} > A ~~for~all~~ n > N$

I've reread he page over and over again but it's not sinking in. I probably just need to sleep on the idea!

2. Originally Posted by aceband
Hey

Can anyone explain what A and N are in the definition for a limit of infinity in a sequence:

$a_{n} > A ~~for~all~~ n > N$

I've reread he page over and over again but it's not sinking in. I probably just need to sleep on the idea!
So, what this definition is saying is that, if $a_n$ repesents a sequence of numbers, then this sequence grows without bound if each member of the sequence is greater than some arbitrarily Large number $A$ whenever $n>N$.

3. Thanks for the quick reply!

I think i struggle to understand because i don't see how it would be used to show an infiinte upper bound (is that correct term).

Where would A and N come from?

4. Originally Posted by aceband

I think i struggle to understand because i don't see how it would be used to show an infiinte upper bound (is that correct term).

Where would A and N come from?
The thing here to realize is that $A$ is chosen to be as large as you wish. So, the proof of an unbounded sequence lies in showing that FOR ANY arbitrarily large numbers $A$ and $N$ (assuming $a_N\geq{A}$), there exists an $n$ such that $a_n>A$ whenever $n>N$

5. Originally Posted by aceband
Hey

Can anyone explain what A and N are in the definition for a limit of infinity in a sequence:

$a_{n} > A ~~for~all~~ n > N$

I've reread he page over and over again but it's not sinking in. I probably just need to sleep on the idea!
A is any finite number. Any number. as big as you want. It is saying that no matter what finite number you pick, all the terms of the sequence will exceed it eventually, after you pass some term $a_n$.

An example, the sequence $\{ a_n \} = n$ goes to infinity as $n$ goes to infinity.

How do we know? well, if some arbitrary number $A$ is given, all we have to do is choose any $N > \lceil A \rceil$ (the smallest integer greater than $A$) and then if $n > N$, $a_n > A$ (that is, all terms after the $N$th term will be bigger than the number $A$.

Example, say someone choose $A = 14.7$, we can pick, $N = 16$, and so $a_{16},~a_{17},~a_{18}, \dots$ will all be bigger than $A$, these will be the integers 16, 17, 18, ...

got it? No matter what number you pick, our sequence will surpass it eventually, in fact, every term of our sequence from some point on will pass it. the fact that this will happen as you keep picking bigger and bigger numbers, means the terms of our sequence can get as big as you want. in technical terms, the sequence tends to infinity

6. Okey doke,

See if i can get this:
Let $a_{n} = n^{2}$

Which should tend to infinity.
To show it i need to show that there exists an n where ' $a_{n}$ > A' and 'n > N'

Is the first step to put n = A?

7. Originally Posted by aceband
Okey doke,

See if i can get this:
Let $a_{n} = n^{2}$

Which should tend to infinity.
To show it i need to show that there exists an n where ' $a_{n}$ > A' and 'n > N'

Is the first step to put n = A?
well, that would work actually--except if 0 < A < 1.

i thought getting the idea was the important thing here. do you get the idea?

to apply it, we have to be a little more technical and stick to the definition closely. you would prove this as follows:

Let $A > 0$ be given. Choose $N = \text{max} \left \{ 1, \sqrt A \right \}$ (this is to deal with the case where 0 < A < 1, if it is in that range, we simply choose N = 1)

Then, $n > N$ implies $a_n = n^2 > N^2 = \sqrt A ^2 = |A| \ge A$.

So $\lim_{n \to \infty} a_n = \infty$ by definition.

8. Hmm i think this is just too much for me to grasp. I can see it hsouldn't be so hard to understand but it just wont click!

9. Originally Posted by aceband
Hmm i think this is just too much for me to grasp. I can see it hsouldn't be so hard to understand but it just wont click!
it's hard for everyone when just learning this. don't be discouraged. how about having a crack at this one...

Show that $\lim_{n \to \infty}(2n + 1) = \infty$

10. Thanks for the words,

Ok lets see then.

Is it along these lines:
$

a_{n} = 2n + 1 \\$

$
a_{n} > A \\$

$
So~~2n+1 > A ~~for~all~~ n > N

$

Before i hash it up any more do i need ot rearrangem get n on it's own at all?

11. Originally Posted by aceband
Thanks for the words,

Ok lets see then.

Is it along these lines:
$

a_{n} = 2n + 1 \\$

$
a_{n} > A \\$

$
So~~2n+1 > A ~~for~all~~ n > N

$

Before i hash it up any more do i need ot rearrangem get n on it's own at all?
you can just solve for n as you would in a regular equation. then you will know what to pick N to be. (By the way, nothing you do so far would show up in your proof. this is just the rough work to know what you'll put into it.

12. So n = 1/2(A - 1) which in turn is greater than N?

13. Originally Posted by aceband
So n = 1/2(A - 1) which in turn is greater than N?
no, we want N to be bigger than that.

think you can plug it into a proof now? follow my example.

14. After several more hours writing out the examples i cant grasp it! The idea of chosing maths as a degree choice is fading!

Can anyone recommend good texts to read that might shed new light for me?

15. Originally Posted by aceband
After several more hours writing out the examples i cant grasp it! The idea of chosing maths as a degree choice is fading!

Can anyone recommend good texts to read that might shed new light for me?
What level do you wish to learn at? Are you doing calculus, analysis, or what?

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