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 $\floor {\frac {n + j b^k} {b^{k + 1} } }$
Let $n, b \in \Z$ such that $n < 0$ and $b \ge 2$.

Then:


 * $\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }= n + 1$

Proof
From the working of Sum over $k$ of Sum over $j$ of $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$:


 * $\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } } = \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } }$

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

and so:
 * $\ds \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } } = n - \paren {-1}$

Hence the result.