Sum of series $\frac{x^k}{n-k}$

Question

Is it possible to find closed form of following sum: $\sum\limits_{k=1}^{n-1}\frac{x^k}{n-k}$ ?
If not then maybe there is at least some good aproximation?

Answer 1

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 $x$.

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

Answer 2

Tomek:

Here is a suggestion, perhaps it is not giving you as explicit a formula as you'd like. Let $f_n(x)$ the sum that you are interested in. Then $$g_n(x):=x^{-n}f_n(x)=\sum_{k=1}^{n-1}\frac{x^{k-n}}{n-k}=\sum_{k=1}^{n-1}\frac{x^{-k}}{k}.$$ Let $y=1/x$, and $h_n(y):=g_n(1/y)$. Differentiate to get $$\sum_{k=1}^{n-1}y^{k-1}=\frac{1-y^{n-1}}{1-y}.$$ This gives you a formula for $h_n(y)$ after integrating (if you try to find it explicitly rather than leaving it in terms of the integral, this formula is a bit messy because it depends on $n$). From this, you easily get the formula for $f_n(x)$.