Finding the Taylor Series of this function I am trying to find a series expansion of the following function:
$$\left(\frac{\log x}{x}\right)^n$$
I need hints or methods for going about doing this. Is it even possible?

I am on to something with the general case without the exponent (With some help form robjohn):
$$D^n\left(\frac{\log(x)}{x}\right)=(-1)^nn!\frac{\log(x)-H_n}{x^{n+1}}$$
 A: If you're curious, I was doing the same over Christmas recently. Let $D^n=\frac{d^n}{dx^n}$.
Define the recursive Harmonic Numbers as follows (with extra conditions for our purposes) for $n,k \in \mathbb{N}_0$ as
$\text{H}_{0,n}=1 \quad\forall \;n\in \mathbb{N}_0$
$\text{H}_{k,n}=0 \quad\text{if } n<k$ 
$\text{H}_{k,n}:=\text{H}_{k,n-1}+1/n\cdot\text{H}_{k-1,n-1} \quad\text{for } n,m\geq1$.
Note that $\text{H}_{1,n}$ are the ordinary harmonic numbers.
Then let $n,m \geq 1$
$D^n \left(\log x\right)^m=\frac{m\left(-1\right)^{n-1}\left(n-1\right)!}{x^n}\left(\sum\limits_{k=1}^m (-1)^{k+1}H_{ k-1,n-1}\; {}^{m-1}P_{m-k} \;\left(\log x\right)^{m-k}\right)$
with ${}^{m-1}P_{m-k}:=\frac{(m-1)!}{(m-k)!}$
Providing there isn't a typo you can verify this by induction. It involves pulling out the first term in a sum, relabeling, combining the two sums back in to one and using the above relation for $\text{H}_{m,n}$.
Maybe there's a modification of this that'll help you. Also not a chance this would fit as a comment. Also also, I under the impression steve roman might have derived this in one of his papers on Harmonic Logarithms but I haven't the chance to read the paper fully.
