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:

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

where $\floor {\, \cdot \,}$ denotes the floor function.

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.

$\blacksquare$

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $44$