Closed form or simplification of $\sum_{k=1}^{n} \frac{x^{n-k}(1-x^k)}{k}$

Question

Does there exist a simpler form of

$$\sum_{k=1}^{n} \dfrac{x^{n-k}(1-x^k)}{k}$$

For $x \in ]0,1[$. Perhaps some combinatorial interpretation?

Answer 1

Rewrite the sum as $$\sum_{k=1}^{n} \dfrac{x^{n-k}(1-x^k)}{k} = \sum_{k=1}^{n} \dfrac{x^{n-k}}{k} - \sum_{k=1}^{n} \dfrac{x^n}{k}= x^n \left[\sum_{k=1}^{n} \dfrac{x^{-k}}{k} - \sum_{k=1}^{n} \dfrac{1}{k}\right]$$

And use the fact that $$\sum_{k=1}^{n} \dfrac{1}{k} = H(n)$$ where $H(n)$ is the Harmonic number.

Also note that $$\sum_{k=1}^{n} \dfrac{x^{-k}}{k} = \left(\frac{1}{x}\right)^{n+1}\left[-\Phi\left(\frac{1}{x},1,n+1\right) \right]-\log \left(\frac{x-1}{x}\right)$$

where $\Phi(x,s,a)$ is called the Lerch transcendent, as per list of known identities.

Answer 2

Expression of the form$\sum_{k=1}^{n} \frac{x^k}{k}$, can be rewritten as $\int_{0}^{x}\sum_{k=1}^{n}y^{k-1}dy$. You can sum (up to $n$), adn then take the integral. Keep in mind $\int \frac{dx}{1-x} = -\log(1-x) +C$