Powers and Power Series' When doing some problems I came across the function:
$$f(x)=\frac{x}{1-2x}$$
I realised that the Maclaurin expansion of this function was:
$$f(x)=x+2x^2+4x^3+16x^4...$$
Evaluate at $x=1$ to get
$$f(1)=1+2+4+16...$$
I have a few of questions about this:
1) By playing around $$g(x)=\frac{x}{1-nx}$$ gives the series for exponents of $n$, is this conjecture true?
2) Is there a way to prove it, any hints?
3) I am aware that the sum of exponents obviously do not converge but why do I get a non-integer (and negative at that) when I evaluate $\frac{x}{1-2x}$ at $x=1$ given that the power series is the sum of integers? 
Can a function only be equal to its corresponding power series if it converges for every value of that function?
Bare in mind that being at secondary school I have not covered analysis of any kind so I am woefully incompetent at such topics.
 A: Generally, given a power series $f(x) = \sum_{k-0}^\infty f_k x^k$, there is a
radius of convergence $R$ associated with the terms $f_k$. Then if $|x|<R$, the
series converges absolutely, if $|x|>R$ then the series diverges and, without
knowing anything more, we can say nothing about convergence for $|x|=R$.
The radius of convergence is given by a formula (see here), in the example
you have above, we can compute
$${1 \over R} = \lim_{n\to \infty} \left| {f_{k+1} \over f_k } \right| = \lim_{n\to \infty}{{1 \over 2} 2^{n+1} \over {1 \over 2} 2^n} = 2$$
And so $R = {1 \over 2}$. Hence the series will diverge at $x=1$.
For 1) For $|x|<1$ we have $1+x+x^2+... = {1 \over 1-x}$, so
$x+nx^2+n^2x^3+... = {1 \over n} (nx+ (nx)^2+ (nx)^3+... )= {1 \over 1-nx}$, and this will be valid for $|x| < {1 \over n}$.
This answers 2).
3) The issue here is that $|x|>R$, so while you have some function $f$ and a power series that equals the function for $|x|<R$, you cannot use the series for $|x|>R$. The function $f$ may be perfectly well defined for $|x|>R$ (as in
your example) but it does not match its power series based at zero.
