Expanding random N(0,1) variable If I have an expression 
$$\frac{1}{1+\sigma m(z/l)}$$
where $m(z/l)$ is a random $N(0,1)$ variable, $\sigma$ is dimensionless, can I rewrite this via an expansion to bring up the random variable on the numerator?
 A: If you want to use $$\frac1{1+x}=1-x+x^2-x^3+\cdots$$ 
then you need to remember that this only works for $|x|\lt 1$
and that a random variable with a normal distribution has a positive probability of being outside this
A: Here are some observations which might be considered as a solution to the problem which was not particularly sharply defined.
Let the random variable in question be
$$x = \frac{1}{1+ \sigma m}\tag{1}$$
where the PDF of $m$ is
$$f(m) = N(0,1) =2 \frac{e^{-\frac{m^2}{2}}}{\sqrt{2 \pi }}\tag{2}$$
and $\sigma \gt 0$.
Notice that we have defined a one-sided PDF ($m\ge 0$) in order to avoid trivial divegencies in $x$.
If you wish to "bring up the variable $m$ into the numerator" you can use the finite expansion as already proposed by Henry:
$$x_s = \sum _{k=0}^{k_{max} } (-1)^k m^k \sigma ^k = 1 - (m \sigma ) + (m \sigma )^2 \mp ... + (- m \sigma )^{k_{max}}\tag{3} $$
The sum cannot be extended to $k_{max} = \infty$ because it is divergent (see below).
That the finite expansion is useful, however, can be seen by studying expectation value of $x$ and hence the moments of $m$.
The $k$-th moment of $x$ is
$$\mu (k) = E[x^k] = \int_{0}^{\infty} m^k f(m)\,dm =\frac{1}{\sqrt{2 \pi } }2^{\frac{k+1}{2}} \Gamma \left(\frac{k+1}{2}\right)\tag{4}$$
The moments increase sharply with $k$.
Taking the expectation value of the expansion $(3)$ we get
$$E[x_s] =  \sum _{k=0}^{k_{max} } (-1)^k \mu (k) \sigma ^k = 1 -\sqrt{\frac{2}{\pi }} \sigma +\sigma ^2 -2 \sqrt{\frac{2}{\pi }} \sigma ^3+3 \sigma ^4 -8 \sqrt{\frac{2}{\pi }} \sigma ^5 +15 \sigma ^6 -48 \sqrt{\frac{2}{\pi }} \sigma ^7+ 105 \sigma ^8...\tag{5}$$
This is not convergent for $k_{max}\to \infty$ for any $\sigma$ as can be verified from $(4)$ e.g. by the quotient criterion.
Calculating the expectation value of $x$ directly gives
$$E[x] =  \int_0^{\infty } \frac{1}{1+m \sigma }f(m) \, dm =\frac{2}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2 \sigma ^2}} \left(\pi \; \text{erfi}\left(\frac{1}{\sqrt{2} \sigma }\right)-\text{Ei}\left(\frac{1}{2 \sigma ^2}\right)\right) \tag{6} $$
This closed expression in terms of the error function and the exponential integral is valid for any $\sigma \gt 0$.
For small $\sigma \gt 0$ this can be expanded into a divergent sum starting as $(5)$. Which justifies the use of $(3)$.
Finally, the PDF $g(x)$ of $x$ can be easily calculated from $f(m) dm = g(x) dx$ i.e. $g(x) = f(m) \frac{dm}{dx}$ and $m = \frac{\frac{1}{x}-1}{\sigma }$ to be
$$g(x) = \frac{\sqrt{\frac{2}{\pi }} e^{-\frac{\left(\frac{1}{x}-1\right)^2}{2 \sigma ^2}}}{\sigma  x^2}\tag{7}$$
for $0 \lt x \le1$ and $g(x) = 0$ elsewhere.
