Element of Leibniz Harmonic Triangle as Sum of Elements on Diagonal from Below

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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

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