Sum over k of Sum over j of Floor of n + jb^k over b^k+1

Theorem
Let $n, b \in \Z$ such that $n \ge 0$ and $b \ge 2$.

Then:


 * $\displaystyle \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor = n$

Proof
We have that $\left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor$ is in the form $\left \lfloor{\dfrac {m k + x} n}\right \rfloor$ so that:

Thus:

Hence the result.