Harmonic triangle

The Harmonic Triangle or Leibniz's Harmonic Triangle by Gottfried Wilhelm Leibniz is structured analogously to Pascal's triangle :

• The nth line begins and ends at the margin ${\displaystyle {\frac {1}{n}}}$
• Each number is the sum of the two numbers below it

${\displaystyle {\begin{array}{cccccccccccccccccc}&&&&&&&&&1&&&&&&&&\\&&&&&&&&{\frac {1}{2}}&&{\frac {1}{2}}&&&&&&&\\&&&&&&&{\frac {1}{3}}&&{\frac {1}{6}}&&{\frac {1}{3}}&&&&&&\\&&&&&&{\frac {1}{4}}&&{\frac {1}{12}}&&{\frac {1}{12}}&&{\frac {1}{4}}&&&&&\\&&&&&{\frac {1}{5}}&&{\frac {1}{20}}&&{\frac {1}{30}}&&{\frac {1}{20}}&&{\frac {1}{5}}&&&&\\&&&&{\frac {1}{6}}&&{\frac {1}{30}}&&{\frac {1}{60}}&&{\frac {1}{60}}&&{\frac {1}{30}}&&{\frac {1}{6}}&&&\\&&&{\frac {1}{7}}&&{\frac {1}{42}}&&{\frac {1}{105}}&&{\frac {1}{140}}&&{\frac {1}{105}}&&{\frac {1}{42}}&&{\frac {1}{7}}&&\\&&{\frac {1}{8}}&&{\frac {1}{56}}&&{\frac {1}{168}}&&{\frac {1}{280}}&&{\frac {1}{280}}&&{\frac {1}{168}}&&{\frac {1}{56}}&&{\frac {1}{8}}&\\&&&&&\vdots &&&&\vdots &&&&\vdots &&&&\\\end{array}}}$

The entries are identified with the symbol , the numbering of the rows and columns starting with 1 (this is not handled uniformly in the literature (starting with 0 or 1)). ${\displaystyle \left[{n \atop k}\right]}$

The recursion applies

${\displaystyle \left[{n \atop 1}\right]=\left[{n \atop n}\right]={\frac {1}{n}},\qquad \left[{n \atop k}\right]=\left[{n+1 \atop k}\right]+\left[{n+1 \atop k+1}\right],\quad n\geq 1,1\leq k\leq n}$

A connection with the binomial coefficients of Pascal's triangle is given by

${\displaystyle \left[{n \atop k}\right]={\frac {1}{k{\binom {n}{k}}}}={\frac {1}{n{\binom {n-1}{k-1}}}}}$, d. H. the entries are fractions .

Because of the sum of the denominators in the nth row . Example: . ${\displaystyle \sum _{k=0}^{n}{\binom {n}{k}}=2^{n}}$ ${\displaystyle n\cdot 2^{n-1}}$ ${\displaystyle 5+20+30+20+5=5\cdot 2^{4}=80}$

The sum of a diagonal results in because

${\displaystyle \left[{n+k \atop k}\right]=\left[{n+k-1 \atop k}\right]-\left[{n+k \atop k+1}\right]}$

the telescope sum

${\displaystyle \sum _{k=1}^{\nu }\left[{n+k \atop k}\right]=\left[{n \atop 1}\right]-\left[{n+\nu \atop \nu +1}\right]={\frac {1}{n}}-\left[{n+\nu \atop \nu +1}\right].}$

Because of the trunk breaks, Leibniz's series follows through the border crossing :

${\displaystyle \sum _{k=1}^{\infty }\left[{n+k \atop k}\right]={\frac {1}{n}}}$for or for ${\displaystyle n\geq 1}$ ${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{\binom {n+k}{k}}}={\frac {n}{n-1}}}$ ${\displaystyle n\geq 2}$

history

In 1672 Christiaan Huygens had given his young friend Leibniz the task of summing the reciprocal triangular numbers:

${\displaystyle 1+{\frac {1}{3}}+{\frac {1}{6}}+{\frac {1}{10}}+\dots +{\frac {2}{n(n+1)}}+\dots }$

He gives 2 as the sum. During his stay in Paris he studied the writings of Blaise Pascal in detail . In a later version of his Historia et Origo , he contrasts Pascal's triangle with its harmonic triangle. The series then results from the general series for n = 2 .

literature

• Joseph Ehrenfried Hofmann , Heinrich Wieleitner , Dietrich Mahnke . The difference calculation at Leibniz . Session reports of the Prussian Academy of Sciences, Physikalisch-Mathematische Klasse, 1931, pp. 562–590. Esp. Pp. 566-572.

Web links

• Eric W. Weisstein : Leibniz Harmonic Triangle . In: MathWorld (English).
• Follow A003506 in OEIS