Here's a very simple yet quite good approximation to $\sum\nolimits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}}$, suitable for $|x|>1$ and $n$ sufficiently large (which is the interesting case, in my opinion).
First write 
$$
S(n,x) := \sum\limits_{k = 1}^{n - 1} {\frac{{x^k }}{{n - k}}} = x^n \sum\limits_{k = 1}^{n - 1} {\frac{{u^k }}{k}},
$$
where $u=1/x$. If $|x|>1$, then $|u|<1$; hence $\sum\nolimits_{k = 1}^\infty  {\frac{{u^k }}{k}}  =  - \log (1 - u)$. 
So, noting that $ - \log (1 - \frac{1}{x}) = \log (\frac{x}{{x - 1}})$, this gives rise to the approximation
$$
S(n,x) \approx x^n \log \bigg(\frac{x}{{x - 1}}\bigg),
$$
provided that $n$ is sufficiently large. It turns out that this simple approximation is quite good. (You can check by yourself.) See the table for an example.

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Rounded values of $S(n,x)$ and its approximation, for $n=10$ and several values of $x$.

$$\begin{array}{ccc}x&S(n,x)&x^n \log\left(\frac{x}{x-1}\right)\\-5&-1780484.04&-1780483.95\\-3&-16987.42&-16987.34\\2&709.598&709.783\\
4&301656.39&301656.52\\6&1.102428722 \times 10^7&1.102428734 \times 10^7\end{array}$$