Expressing $e^{-x}$ through a different series I'm consider $e^{-x}$ with $x >> 0$. I want to express it in a series of the form
$$ e^{-x} = \sum_{i=1}^{\infty} \frac{a_i}{x^k}$$
But analysis seems to resist it:
Motivation:
$ \frac{1}{x^k}$ rapidly decays to 0, as does $e^{-x}$ so I'm thinking a sum of such terms should intuitively yield a series.
Work So Far:
My first idea was that substitution $ u = \frac{1}{x}$ yield that 
$$ e^{-x} = e^{-\frac{1}{u}}$$ 
Now I could look to express this as a Taylor series around $u = \gamma$, observing that 
$$ \frac{d}{du} e^{ - \frac{1}{u}} = \frac{1}{u^2} e^{-\frac{1}{u}}  $$
Deriving again we  have 
$$  \frac{d^2}{du^2} e^{ - \frac{1}{u}} = \left(- \frac{2}{u^3} + \frac{1}{u^4} \right)e^{-\frac{1}{u}} $$
Again:
$$  \frac{d^3}{du^3} e^{ - \frac{1}{u}} = \left( \frac{2\times 3}{u^4} - \frac{4}{u^5} + \frac{1}{u^6}\right)e^{-\frac{1}{u}} $$
And In General
$$ \frac{d^k}{du^k} e^{ - \frac{1}{u}}= $$
$$ \left((-1)^{k-1} \frac{2 \times 3 \times ... k}{u^{k+1}} + (-1)^{k-2} \frac{4 \times 5 \times 6 ... \times (k+1)}{u^{k+2}}  + (-1)^{k-3} \frac{6 \times 7 \times ... (k+2)}{u^{k+3}}... +\frac{1}{u^{2k}} \right) e^{-\frac{1}{u}} $$
$$ = e^{- \frac{1}{u}} \sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} \frac{(-1)^{-i}}{u^{2k-i}} = \frac{e^{-\frac{1}{u}}}{u^{2k}}\sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} (-1)^{i} u^{i} $$
Then it follows that we can define $e^{-\frac{1}{u}}$ around a point $u = \gamma$ as 
$$ e^{- \frac{1}{u}} = e^{-\frac{1}{\gamma}} + \sum_{k=1}^{\infty} \left(\frac{e^{-\frac{1}{\gamma}}}{k!\gamma^{2k}}\sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} (-1)^{i} \gamma^{i} \right) ( u - \gamma)^{k}   $$ 
And therefore:
$$ e^{-x} =  e^{-\frac{1}{\gamma}} + \sum_{k=1}^{\infty} \left(\frac{e^{-\frac{1}{\gamma}}}{k!\gamma^{2k}}\sum_{i = 0}^{k-1} \frac{(2k-1-i)!}{(2k -1 -2i)!} (-1)^{i} \gamma^{i} \right) \left(  \frac{1}{x} - \gamma \right)^{k}  $$
But this explodes if $\gamma = 0$, so the mechanics of building THAT converging series are going to be very different than this (namely NO convergence around $x < 1$ and then something nice...)
Error:
There is an error in my derivation, namely: 
$$  \frac{d^3}{du^3} e^{ - \frac{1}{u}} = \left( \frac{2\times 3}{u^4} - \frac{6}{u^5} + \frac{1}{u^6}\right)e^{-\frac{1}{u}} $$.
The remaining series derivation is also incorrect for the same mistake not caught.
 A: $$\frac{\partial}{\partial x} \exp(-x)=-\exp(-x)\\
\frac{\partial}{\partial x}\sum_{i=1}^{\infty} a_i x^{-i}=\sum_{i=1}^{\infty} \frac{a_i}{-i} x^{-i-1}$$
We would like
$$\sum_{i=1}^{\infty} a_i x^{-i}=\sum_{i=1}^{\infty} \frac{a_i}{i} x^{-i-1}$$
If the coefficients in $x^{-1}$ are equal, then $a_1$ must be 0, because the rhs. has no term in $x^{-1}$. By induction we see that all $a_i$ must be 0. Thus, there is no such series that approximates $\exp(-x)$.
You may be interested in Pade approximations.
A: You wish to have an asymptotic expansion of $e^{-x}$ as $x \to \infty$. Allowing $x$ to be complex, and writing $z = 1/x$, we have $e^{-1/z}$ for which we want a Taylor series expansion around $z = 0$. Like Wouter has pointed, this is impossible because $e^{-1/z}$ has an essential singularity at $z = 0$. You may write a Laurent series, but I expect this to be identical to the Taylor series of $e^{-x}$.
This also shows that $e^{-x}$ is a term that is "beyond all orders" when it comes to asymptotics. It decays faster than every possible power law.
