While we're at it, let's suppose you have the following problem.

You have a function, but you really like polynomials. You wish the function was a polynomial instead. So approximate it by one, but which one? Well, it depends on what kind of shape you want. Maybe you only need the approximation near a point (say x=0). So what to do? Well, make sure the functions match. But the shape should stay the same, too. So the derivatives should match as well. What degree of polynomial? Well, however many derivatives you want to match. If you want to match all the derivatives, then you need a polynomial of all the degrees (i.e., a series).

How do you match up a series with a function such that all the derivatives agree at a single point? Well do the obvious calculation, and you get the Taylor series, and low and behold the Taylor series for 1/(1-x) is just 1+x+x^2+...

I'll put details when I'm back from phone posting.\

Edit:

Let f:R->R be a smooth (infinitely differentiable) function and fix some x0 in R. Consider the power series

F(x) = a_0+a_1 (x-x0)+ a_2 (x-x0)^2 + ... + a_n(x-x0)^n + ...

We want F to be the "best" approximation to f "near" x0, in the sense that F agrees with f at x0 in function value and in all derivatives of arbitrary order. Then we simply require:

F(x0) = f(x0)

F'(x0) = f'(x0)

F''(x0) = f''(x0)

...

F^[n](x0) = f^[n](x0), (this means nth derivative)

...

Well, what is F(x0)? It's just

F(x0) = a_0 + a_1(x0-x0)+ a_2(x0-x0)^2 + .... = a_0 + 0 = a_0

What is F'(x0)?

F'(x) = a_1 + 2a_2(x-x0) + 3a_3(x-x0)^2 + ... + na_n (x-x0)^(n-1) + ...

F'(x0) = a_1 + 2a_2(x0-x0) + 3a_3(x0-x0)^2 + ... + na_n (x0-x0)^(n-1) + ... = a_1 + 0 = a_1

Continuing,

F''(x) = 2a_2 + 6a_3(x-x0) + ... + n(n-1)a_n (x-x0)^(n-2) + ...

F''(x0) = 2a_2+ 0 + 0 + ... = 2a_2

And in general

F^[n](x) = n! a_n + ((n+1)!/1!) a_{n+1}(x-x0) + ((n+2)!/2!) a_{n+2}(x-x0)^2 + ...

so that

F^[n](x0) = n! a_n.

Then we simply require

F(x0) = a_0 = f(x0)

F'(x0) = a_1 = f'(x0)

F''(x0) = 2a_2 = f''(x0)

F'''(x0) = 6a_3 = f'''(x0)

...

F^[n](x0) = n!a_n = f^[n](x0)

...

So in general,

a_n = f^[n](x0)/n!.

Then the series F is

F(x) = \sum_{n=0}^\infty (f^[n](x0)/n!) (x-x0)^n

Now, when f(x) = 1/(1-x) and x0 = 0, what are the derivatives? Well,

f(0) = 1

f'(x)= -(1-x)^(-2)*(-1) = (1-x)^(-2), f'(0) = 1

f''(x) = 2(1-x)^(-3), f''(0) = 2

f'''(x) = 6 (1-x)^(-4), f'''(0) = 6,

...

pattern is you multiply by 1,2,3,4, ...., etc. (every derivative has an extra -1 from the chain rule applied to (1-x)), so you just end up with

f^[n](x) = n! (1-x)^(-(n+1)),

so

f^[n](0) = n!

Then

a_n = (f^[n](x0)/n!) = n!/n! = 1,

so

F(x) = \sum_{n=0}^\infty x^n

is the Taylor series approximation of 1/(1-x).

Then you can prove that indeed convergence occurs (and is uniform) for \abs{x} < 1.