Element of Leibniz Harmonic Triangle as Sum of Elements on Diagonal from Below
Theorem
Consider the Leibniz harmonic triangle:
- $\begin{array}{r|rrrrrr}
n & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & \frac 1 1 \\ 1 & \frac 1 2 & \frac 1 2 \\ 2 & \frac 1 3 & \frac 1 6 & \frac 1 3 \\ 3 & \frac 1 4 & \frac 1 {12} & \frac 1 {12} & \frac 1 4 \\ 4 & \frac 1 5 & \frac 1 {20} & \frac 1 {30} & \frac 1 {20} & \frac 1 5 \\ 5 & \frac 1 6 & \frac 1 {30} & \frac 1 {60} & \frac 1 {60} & \frac 1 {30} & \frac 1 6 \\ \end{array}$
Let $\tuple {n, m}$ be the element in the $n$th row and $m$th column.
Then:
- $\tuple {n, m} = \ds \sum_{k \mathop \ge 0} \tuple {n + 1 + k, m + k}$
Lemma 1
- $\tuple {n, m} = \tuple {n + 1, m} + \tuple {n + 1, m + 1}$
That is, each number in the Leibniz harmonic triangle is equal to the sum of the number below it and the number to the right of that number.
Lemma 2
- $\ds \forall r \in \N_{>0}: \tuple {n, m} = \tuple {n + r, m + r} + \sum_{k \mathop = 1}^r \tuple {n + k, m + k - 1}$
That is, each number in the Leibniz harmonic triangle is equal to the sum of the number below it, $\paren {r - 1}$ numbers diagonally below that number, and the number to the right of the last number.
Proof
Taking $r \to \infty$ in Lemma 2:
\(\ds \sum_{k \mathop \ge 0} \tuple {n + 1 + k, m + k}\) | \(=\) | \(\ds \sum_{k \mathop \ge 1} \tuple {n + k, m + k - 1}\) | Translation of Index Variable of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{r \to \infty} \paren {\tuple {n, m} - \tuple {n + r, m + r} }\) | Lemma 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {n, m} - \lim_{r \to \infty} \frac 1 {\paren {n + r + 1} \binom {n + r}{m + r} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {n, m}\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $35$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $35$