An asymptotic expansion for $(1 + \frac{x}{n})^n$.

Question

I am trying to work out an asymptotic expansion for the function $$f(x, n) = \left(1 + \frac{x}{n}\right)^n$$ in the following sense. For all $k \geq 1$, let $f_k(x)$ be the function recursively defined by $$f_k(x) = \lim_{n\to\infty} n^k \left(f(x, n) - \sum_{j = 0}^{k-1}\frac{f_j(x)}{n^j}\right)$$ where $f_0(x) = e^x$.

My plan of attack is write out $f(x, n)$ and rearrange terms to pull out factors of $\frac{1}{n^j}$. Hopefully the coefficients of these powers will be easily identifiable power series. Instead of remaking the wheel though, is this a well known expansion? Are there any suggestions for how to make this an easier task?

Thanks.

Answer 1

Okay, I have a couple of unpleasant ways to offer; someone else may be able to improve on them.

1. Start with $e^{-x}(1+x/n)^n$, and do a series expansion about $x=0$. This will give you a series where you get powers of $n$ in the denominator, but more importantly, you only have to go to the $2k$th term to get the full expansion for the $k$th power of $n$. (There should be a simple proof of this, but for the moment it eludes me.)
2. Take the logarithm, subtract $x$, and use the power series $$ n\log{(1+x/n)}-x = \sum_{k=2}^{\infty} (-1)^{k-1} \frac{x^k}{kn^{k-1}} $$ and now use the exponential formula to recover the power series of the exponential. This requires that you look up the Bell polynomials, but from there it is straightforward and algorithmic, if not necessarily as closed a form as you would like.