Infinite sum of floor functions I need to compute this (convergent) sum
$$\sum_{j=0}^\infty\left(j-2^k\left\lfloor\frac{j}{2^k}\right\rfloor\right)(1-\alpha)^j\alpha$$
But I have no idea how to get rid of the floor thing. I thought about some variable substitution, but it didn't take me anywhere.

Hint: break the sum up into 
$$\sum_{j=0}^\infty j(1-\alpha)^j\alpha + \sum_{j=0}^\infty\left(-2^k\left\lfloor\frac{j}{2^k}\right\rfloor\right)(1-\alpha)^j\alpha$$
The sum on the left is easy to compute, and you can split the sum on the right up into the regions where $\lfloor\frac{j}{2^k}\rfloor$ takes different values.
For $0\leq j\leq2^k-1$, $\lfloor\frac{j}{2^k}\rfloor = 0$. For $2^k \leq j \leq 2^k+(2^k-1) = 2\times2^k - 1$, $\lfloor\frac{j}{2^k}\rfloor = 1$. Similarly, for any positive integer $m$, for $m2^k \leq j \leq (m+1)2^k -1$, $\lfloor\frac{j}{2^k}\rfloor = m$.
Using this we can split the left sum into 
$$-\alpha2^k\sum_{m=0}^\infty \sum_{j=m2^k}^{(m+1)2^k-1}m(1-\alpha)^j = -\alpha2^k\sum_{m=0}^\infty m(1-\alpha)^{m2^k} \sum_{j=0}^{2^k-1}(1-\alpha)^j $$
$$=-\alpha2^k\sum_{m=0}^\infty m(1-\alpha)^{m2^k} \frac{(1-\alpha)^{2k}-1}{-\alpha} =2^k((1-\alpha)^{2k}-1)\sum_{m=0}^\infty m((1-\alpha)^{2k})^m$$
$$=\frac{2^k((1-\alpha)^{2k}-1)(1-\alpha)^{2k}}{((1-\alpha)^{2k}-1)^2} = \frac{-2^k(1-\alpha)^{2k}}{(1-\alpha)^{2k}-1}$$
( I think!)
