Hi Mattofix,
I'd like to split my answer into three parts.
PART 1
Mattofix said:
I am currently reading through my notes and found the convergent series to be defined.
lxn - Ll is less than (epsilon) whenever n is greater or equal to N
...i have looked on wikpedia and a few other web sites and i am not making any sense of what this N is...
I think we should review what
convergence means (it doesn't matter if
we consider a sequence or a series)
My professor once explained convergence this way which I found good:
Prof: "In order to understand the definition of convergence -
the sequence x_n is convergent to L - let's play a game:
You, dear students, give me an epsilon, and I will give you an N such
that | x_n - L | < \epsilon for all n\geq N."
So, the game consists of:
Step 1: Students choose epsilon.
Step 2: Prof chooses N
Step 3: Prof has to check if following condition is fulfilled:
| x_n - L | < \epsilon if n \geq N
or: if n \geq N then |x_n-L| < \epsilon
---
Mattofix, suppose we play that game then it goes like this:
Let's take e.g. the sequence x_n = 1/n for which I claim
that it converges to L=0.
Now, give me an \epsilon.
Step 1:
Suppose you gave me \epsilon = 0.002.
Then my task is to find an N such that
| x_n - L | < \epsilon or
| 1/n - 0 | < 0.002
if n \geq N.
Step 2:
So, it's my turn to choose an N.
For \epsilon = 0.002 I choose N = 1000.
Does this N fulfill the condition
| x_n - L | < \epsilon or
| 1/n - 0 | < 0.002
if n \geq N?
Step 3:
Let's check if the condition is fulfilled:
From the definition, n shall be greater or equal to N, thus:
n \geq N or n \geq 1000
Thus, from n \geq 1000 we get
1/n \leq 0.001 < 0.002 = \epsilon or
|1/n - 0| < \epsilon
(condition fulfilled)
---
We can play a new round in which you give me another \epsilon.
Then I find an N again for which
| 1/n - 0 | < \epsilon
---------------------------------------------------------------------------------
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PART 2
Mattofix said:
What is this large integer N and why does n have to be greater or equal to it?
Maybe to understand more intuitively why n \geq N (and not only n=N) do the following:
Write down the values of the sequence x_n if n \geq 1000.
Let's keep \epsilon = 0.002.
x_{1000} = 1/1000
x_{1001} = 1/1001
x_{1002} = 1/1002
...
and so on.
What do you notice?
All these values of x_n, namely if n \geq N or n \geq 1000 fulfill the condition
| x_n - L | < \epsilon
| 1/n - 0| < 0.002
To make it clearer, let's write down again the list from above,
but this time we check, whether the condition | 1/n - 0| < 0.002 is fulfilled:
Left hand side: x_n=1/n, right hand side: Check if |1/n - 0| < 0.002
x_{1000} = 1/1000 Check: |1/1000 - 0| < 0.002 (fulfilled)
x_{1001} = 1/1001 Check: |1/1001 - 0| < 0.002 (fulfilled)
x_{1002} = 1/1002 Check: |1/1002 - 0| < 0.002 (fulfilled)
...
and so on
You see, we have chosen an N such that for all n \geq N
the following holds: |x_n - L| < \epsilon
(Note: We've only found an N for epsilon=0.002. If you want to show
that x_n is convergent you have to show that
for any arbitrary epsilon>0 it is possible to find an N.)
You can make it even clearer if you interpret |x_n - L| < \epsilon graphically.
Just plot the sequence x_n.
Have a look at figure 2.2 at the bottom of http://www.maths.abdn.ac.uk/~igc/tch/ma2001/notes/node18.html. In figure 2.2 a sequence is shown together with a blue stripe called the "epsilon neighborhood". You can recognize that all the dots lie
within the epsilon neighborhood after some value N, i.e. if n \geq N.
That is the visual interpretation of:
if n \geq N then |x_n-L| < \epsilon-----------------------------------------------------------------------------------
-----------------------------------------------------------------------------------
PART 3
Mattofix said:
so to confirm...
N is the value of n where lxn - Ll = epsilon
Mattofix, you mentioned: if n = N then |x_n-L|= \epsilon
(contrary to the definition: if n \geq N then |x_n-L| < \epsilon)
This may work sometimes, sometimes not.
Let's see what happens if we use your definition: if n = N then |x_n-L|= \epsilon
Let x_n = 1/n, thus from |x_n-L|= \epsilon we get
|1/n - 0| = \epsilon
a) Example where it works:
Choose \epsilon = 0.01
=> |1/n - 0| = 0.01
Then it's easy to find an N with n=N, namely N = 100:
|1/100 - 0| = 0.01
b) Example where it doesn't work:
Choose \epsilon = 2
=> |1/n - 0| = 2
Then it's not possible to find an N with n=N such
that |1/n - 0|=2.
(Do you see why?)As an exercise, try to show that with the correct definition of
convergence - |x_n -L| < \epsilon if n \geq N - we can choose \epsilon = 2
without getting into trouble, i.e. it is possible to find an N such that
|1/n - 0| < 2 if n \geq N.
