# I can't understand this in partial fractions

ya agreed.

arildno
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But, consider that -1<x<1, that is, |x|<1
What happens to the value of Sn as N grows bigger and bigger?

What does |x|<1 this mean? i mean by using |x|

The value gets smaller and smaller right?

Either positive or negative values right?

arildno
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|x| means the ABSOLUTE value of the number x, and if x is non-zero, the absolute value is always positive.

So the absolute value of "2" is 2, and the absolute value of "-2" is also 2.
The absolute value is simply the distance of a number on the number line from the origin.

So because of the absolute value are we getting positive values for X?

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arildno
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Hmm..not sure what you mean:
The term $x^{N}$ will approach zero as N towards to infinity if |x|<1.
Do you agree to that?

I think I am a little confused. Does this happen because |x| < 1 will at a certain stage reach 0 I mean like x = 0

arildno
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No, think of x as a FIXED number, lying between -1 and 1.
If you multiply a positive number that is less than one with itself, will the product be less than or greater than the number itself?

Well it say the number = y

and |y|<1

then the product will be less than the number right?

I mean when you multiply it by itself.

arildno
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Right!
So when you multiply itself with itself N times, where N is some big number, then $x^{N}$ will be very close to zero,right?

Right right it will be very close to zero. I mean it will go on like 0.0000000000001323 like that right?

arildno
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You've got it.

So, if we have $S_{N}=\frac{1-x^{N}}{1-x}$, |x|<1 and N is really big, what will $S_{N}$ be approximately equal to?

Will Sn approximately be equal to = 1/1-x

Am I right?

arildno
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Perfectly! (that is 1/(1-x), remember parentheses..)

Thus, it gives perfect meaning to say that as N goes to infinity, Sn converges to a number S=1/(1-x), or that the INFINITE series S is a meaningful concept. Agreed?

uhh arildno does converge means like bringing it to one place like? Sorry to ask this because I am from a non-english country?

Ya I agree that now the S has a meaningful value.

Oh please continue please? I can realy understand what you teach me than my school teacher.

arildno
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So, what we have found out, is that the INFINITE series,
$$S=\sum_{n=0}^{\infty}x^{n}$$
is a meaningful concept, as long as |x|<1.
We call 1 here to be the radius of convergence for the infinite series, that is the bound we must put on x, in order for the infinite series to have any meaning (I.e, being some number).
Okay?

Ya right. 1 is the bound because if it gose beyond 1 then we wont find a bound because we can get large numbers right?

Anyway this is the first time I used this symbol next to the "Sn ="
$$S=\sum_{n=0}^{\infty}x^{n}$$

VietDao29
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dilan said:
Ya right. 1 is the bound because if it gose beyond 1 then we wont find a bound because we can get large numbers right?

Anyway this is the first time I used this symbol next to the "Sn ="
$$S=\sum_{n=0}^{\infty}x^{n}$$
$$\sum_{i = m} ^ n$$, this is the summation symbol. It's a capital Sigma.
i represents the index of summation; m is the lower bound of summation, and n is the upper bound of summation.
----------
$$\sum_{i = m} ^ n (a_i)$$
means that all you need is just to sum from am to an
$$\sum_{i = m} ^ n (a_i) = a_m + a_{m + 1} + a_{m + 2} + ... + a_{n - 1} + a_n$$
----------------
Example:
$$\sum_{k = 1} ^ 3 \left( \frac{k}{k + 1} \right) = \frac{1}{1 + 1} + \frac{2}{2 + 1} + \frac{3}{3 + 1} = \frac{23}{12}$$
----------------
For Convergent series, you may want to have a look here: Convergent series.
If you say some series $$S_N = \sum_{k = 1} ^ N (a_k)$$ converges to some number L, then it means that: $$\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}$$
Can you get this? :)

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Ok I got it now

arildno said:
So, what we have found out, is that the INFINITE series,
$$S=\sum_{n=0}^{\infty}x^{n}$$
is a meaningful concept, as long as |x|<1.
We call 1 here to be the radius of convergence for the infinite series, that is the bound we must put on x, in order for the infinite series to have any meaning (I.e, being some number).
Okay?

Hi VietDao29 thanks for your post about the meaning of the Capital letter of zigma with the other letters. Realy useful. Thanks alot.

Ya now I undertand. Okay we have now got a meaningful concept for Sn. Earlier I had a little problem in undertanding the symbol, but now I get it. Okay now we have a meaningful concept as long as |x|<1

$$\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}$$

Anyway arildno I hope that this will not be needed for now. If yes I would like to know about that also.
Okay upto here now I can get it. What's the next step?

VietDao29
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dilan said:
$$\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}$$
Have you learnt limit by the way? Something looks like:
$$\lim_{x \rightarrow 3} x ^ 2 = 9$$ (this is an example of limit of a function)
or:
$$\lim_{n \rightarrow \infty} \frac{1}{n} = 0$$ (an example of limit of a sequence)???

arildno
Okay, just substitute S for L, and the statement $\lim_{N\to\infty}S_{N}=S$ means that the value of $S_{N}$ approaches S as N trundles off into infinity. (Something you already know).
$N\in\mathbb{N}$ just means that N is a natural number (1, 2, 350000)and so on (not a fraction or decimal number).