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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then:

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

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


Corollary

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

We have that $\floor {\dfrac {n + j b^k} {b^{k + 1} } }$ is in the form $\floor {\dfrac {m k + x} n}$ so that:

\(\ds \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }\) \(=\) \(\ds \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b}\)
\(\ds \) \(=\) \(\ds \sum_{0 \mathop \le j \mathop < b} \floor {\dfrac {j + \frac n {b^k} } b} - \floor {\dfrac n {b^{k + 1} } }\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {1 - 1} \paren {b - 1} } 2 + \dfrac {\paren {1 - 1} } 2 + 1 \floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } }\) Summation over k of Floor of mk+x over n
\(\ds \) \(=\) \(\ds \floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } }\)

Thus:

\(\ds \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \floor {\dfrac {n + j b^k} {b^{k + 1} } }\) \(=\) \(\ds \sum_{k \mathop \ge 0} \paren {\floor {\dfrac n {b^k} } - \floor {\dfrac n {b^{k + 1} } } }\)
\(\ds \) \(=\) \(\ds \lim_{k \mathop \to \infty} \floor {\dfrac n 1} - \floor {\dfrac n {b^{k + 1} } }\) Definition of Telescoping Series
\(\ds \) \(=\) \(\ds n\)

Hence the result.

$\blacksquare$


Sources