Element of Leibniz Harmonic Triangle is Sum of Numbers Below

Theorem

The elements in the Leibniz harmonic triangle are the sum of the elements immediately below them.

Proof

By definition of Leibniz harmonic triangle, element $\tuple {n, m}$ is:

$\dfrac 1 {\paren {n + 1} \binom n m}$

Thus we have:

 $\ds$  $\ds \dfrac 1 {\paren {n + 2} \binom {n + 1} m} + \dfrac 1 {\paren {n + 2} \binom {n + 1} {m + 1} }$ Elements of Leibniz Harmonic Triangle immediately below $\ds$ $=$ $\ds \dfrac {\binom {n + 1} {m + 1} + \binom {n + 1} m} {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }$ $\ds$ $=$ $\ds \dfrac {\binom {n + 2} {m + 1} } {\paren {n + 2} \binom {n + 1} m \binom {n + 1} {m + 1} }$ Pascal's Rule $\ds$ $=$ $\ds \dfrac {\frac {\paren {n + 2}!} {\paren {m + 1}! \paren {n - m + 1}!} } {\paren {n + 2} \frac {\paren {n + 1}!} {m! \paren {n - m + 1}!} \frac {\paren {n + 1}!} {\paren {m + 1}! \paren {n - m}!} }$ Definition of Binomial Coefficient $\ds$ $=$ $\ds \dfrac {\paren {n + 1}!} {\frac {\paren {n + 1}!} {m!} \frac {\paren {n + 1}!} {\paren {n - m}!} }$ some simplification $\ds$ $=$ $\ds \dfrac 1 {\frac {\paren {n + 1}!} {m! \paren {n - m}!} }$ further simplification $\ds$ $=$ $\ds \dfrac 1 {\paren {n + 1} \frac {n!} {m! \paren {n - m}!} }$ further simplification $\ds$ $=$ $\ds \dfrac 1 {\paren {n + 1} \binom n m}$ Definition of Binomial Coefficient

Hence the result.

$\blacksquare$