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

Corollary to Sum over $k$ of Sum over $j$ of $\left \lfloor{\frac {n + j b^k} {b^{k + 1} } }\right \rfloor$
Let $n, b \in \Z$ such that $n < 0$ and $b \ge 2$.

Then:


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

Proof
From the working of Sum over $k$ of Sum over $j$ of $\left \lfloor{\frac {n + j b^k} {b^{k + 1} } }\right \rfloor$:


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

But:
 * $\forall k \in \Z_{> 0}: \dfrac n {b^{k + 1} } < 0$

and so:
 * $\displaystyle \lim_{k \mathop \to \infty} \left \lfloor{\dfrac n 1}\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor = n - \left({-1}\right)$

Hence the result.