So I am trying to find the asymptotic behavior of:
$$\left(1+\frac{x}{n}\right)^n$$
For very large $n$. I know I can use $z=\frac{1}{n}$ and obtain:
$$\left(1+xz\right)^\frac{1}{z}$$
And find a laurent series about $z=0$, but my question is how? How do I expand this about $z=0$? Mathematica gives the following result near $z=0$:
$$(1+xz)^\frac{1}{z}\approx e^x -\frac{1}{2}e^x (zx^2)+\frac{1}{24}z^2e^xx^3(3x+8)+...$$
Which is good enough since I only care about the those first two terms for the problem I'm working, but how do I get those terms? Can I get them from the bionomial theorem? I tried messing with that a bit but it didn't seem to lead me anywhere.
$$\ln \left(1+xz\right)^{1/z} = \frac1z \ln(1+xz) = \frac1z \left(xz - \frac12 (xz)^2 + O(z^3)\right) = x - \frac12 x^2z +O(z^2)$$ So: $$(1+xz)^{1/z} = e^x e^{-\frac12x^2z+O(z^2)} = e^x \left( 1 - \frac12 x^2z + O(z^2) \right) = e^x - \frac12 e^x (zx^2) + O(z^2)$$