Evaluating series sum with floor

Question

Is there a closed form expression or a good approximation for the following expression: \begin{equation} \sum^{\infty}_{k=n+1} {\left\lfloor \frac{k}{n+1} \right\rfloor p^k} \end{equation}

Knowing that $n\geq1$ and $0< p< 1$.

Answer 1

Writing $k=(n+1)q+r$, $$\sum_{k=n+1}^\infty\left\lfloor\frac k{n+1}\right\rfloor p^k=\sum_{q=1}^\infty\sum_{r=0}^nqp^{(n+1)q+r}= \sum_{q=1}^\infty qp^{(n+1)q}\cdot \sum_{r=0}^np^r.$$ Now $$\sum_{r=0}^np^r=\frac{1-p^{n+1}}{1-p}$$ and $$ \sum_{q=1}^\infty qx^q=\frac x{(1-x)^2},$$ so $$\sum_{k=n+1}^\infty\left\lfloor\frac k{n+1}\right\rfloor p^k=\frac{1-p^{n+1}}{1-p}\cdot\frac{p^{n+1}}{(1-p^{n+1})^2}=\frac{p^{n+1}}{(1-p)(1-p^{n+1})}. $$