Estimation of geometric sum

Question

Is there a way to estimate the

$\sum_{k=0}^{\infty}kq^{k}(1-q)$ , where $q=1-p$

Can we estimate it as a geometric sum ??

Because it seems that we can but we have that $k$ in front of $q^{k}$ that make things difficult.

Answer 1

If it means that $p$ is a probability and $0<p\leq1$ then we obtain $$(1-q)q\sum_{k=0}^{+\infty}kq^{k-1}=(1-q)q\left(\sum_{k=0}^{+\infty}q^{k}\right)'=q(1-q)\left(\frac{1}{1-q}\right)'=\frac{q}{1-q}$$

Answer 2

for the finite sum we get $$\sum_{k=0}^nkq^k(1-q)=-{\frac {{q}^{n+1} \left( \left( n+1 \right) q-n-1-q \right) }{-1+q}} -{\frac {q}{-1+q}} $$ and take the Limit for $n$ tends to infinity

Answer 3

Note that $$ \frac{1}{(1-q)^2}=\left(\sum_{k=0}^{\infty}q^{k}\right)^2=(1+q+q^2+\dotsb)(1+q+q^2+\dotsb)=\sum_{k=0}^{\infty}(k+1)q^{k}\tag{1} $$ by the cauchy product of series. Hence $$ \frac{q}{(1-q)^2}=\sum_{k=0}^{\infty}kq^{k}.\tag{2} $$ Finally $$ \sum_{k=0}^{\infty}k(1-q)q^{k}=\frac{q(1-q)}{(1-q)^2}=\frac{q}{1-q}\tag{3}. $$