# Convergent series - What is this N thing?

1. Jan 6, 2008

### Mattofix

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...

wikipedia says - 'a series converges if there exists a limit L such that for any arbitrarily small positive number epsilon > 0, there is a large integer N such that for all n greather equal to N.'

I know that epsilson is the measure of closeness so that as n tends to infinity epsilon tends to 0.

What is this large interger N and why does n have to be greater or equal to it?

2. Jan 6, 2008

### morphism

N is used to signify how far into the sequence you have to go to make sure its terms get within an epsilon of the limit. That is, if n >= N, then x_n will be within epsilon of L. If you think about it intuitively, the smaller the epsilon, the longer you have to wait until the tail of the sequence is within an epsilon of the limit.

Here's an example. Take the sequence (x_n) given by x_n = 1/n, whose limit we know to be 0. If we would like (x_n) to be within 1/10 of 0, then we choose N=11, because if n>=11, then |x_n - 0| < 1/10. Of course we could also have chosen N=132974234, or any integer larger than 11.

Last edited: Jan 6, 2008
3. Jan 6, 2008

### belliott4488

N is not a magic number - you should just read those definitions as saying, "... we can find some integer less than infinity, which we'll call N, such that ...."

4. Jan 6, 2008

### Mattofix

so to confirm....

N is the value of n where lxn - Ll = epsilon

...therefore for the example that morphism gave N should/could equal 10 because if n>N, then lxn - 0l<1/10?

5. Jan 6, 2008

### HallsofIvy

Staff Emeritus
First, let me clarify that you are talking about sequences not series.

No, N could not equal N because the definition you gave, from Wikipedia, was "n is greater or equal to N". With n= 10, |1/n|= 1/10= $\epsilon$, not "<". As morphism said, N must be larger than n. If you require that N be an integer (not all texts do) then N must be "larger than or equal to 11". but, in any case, certainly must be larger than 10.

6. Jan 6, 2008

### Mattofix

http://en.wikipedia.org/wiki/Convergent_series

...what i am saying is that when/if n=N then lxn-Ll = epsilon, which if correct would answer my question of what is N?

But as you said by the definition, n>N and lxn-Ll<epsilon which is the requirements for a convergent series.

...i think...

7. Jan 7, 2008

### HallsofIvy

Staff Emeritus
Then why don't you ask a question about series? If $|x_n- L|< \epsilon$ then L is the limit of the sequence xn, not of the series $\sum_{n=1}^\infty x_n$.

I'm not at all sure what you mean by that. If, for any $\epsilon> 0$, there exist N such that whenever n> N $|x_n- L|< \epsilon$ then the sequence converges to L. That is not at all what you said.

8. Jan 7, 2008

### Edgardo

Hi Mattofix,

I'd like to split my answer into three parts.

PART 1

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

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, 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 $n \geq N$.

Last edited by a moderator: Apr 23, 2017