1. Leibniz’s Harmonic Triangle Gottfried Leibniz, 1762 Stephanie Moore & Jennifer LeBlanc

2. Background Gottfried Leibniz Blaise Pascal Yang Hui Leonhard Euler
History of Triangles
Sanskrit poet Pingala’s Mythical mountain - 3rd Century
Yang Hui - 13th Century
Pascal - 17th Century
Euler – 18th Century
Leibniz – 18th Century
Leibniz formed his triangle in response to a challenge from Christiaan Huygens to find the sum of the triangle reciprocals.
Gottfried Leibniz
Blaise Pascal
Yang Hui
Leonhard Euler

3. Leibniz’s Harmonic Triangle
Pascal’s Triangle
Leibniz’s Harmonic Triangle 1 1
𝟏 𝟐 1 𝟏
𝟏 𝟐 1
𝟏 𝟑 2
𝟏 𝟔 1
𝟏 𝟑 1
𝟏 𝟒
𝟏 𝟏𝟐 3 3
𝟏 𝟏𝟐 1
𝟏 𝟒
Pascal’s Triangle – ADD DOWN, coefficients for binomial expansion, Combinations
Leibniz Triangle – ADD UP, he created it in response to Christiaan Huygen’s challenge: Sum of triangle reciprocals. 1
𝟏 𝟓 4
𝟏 𝟐𝟎 6
𝟏 𝟑𝟎 4
𝟏 𝟐𝟎 1
𝟏 𝟓 1
𝟏 𝟔 5
𝟏 𝟑𝟎 10
𝟏 𝟔𝟎 10
𝟏 𝟔𝟎 5
𝟏 𝟑𝟎 1
𝟏 𝟔

4. Relationships Between the Triangles1 𝟏 2 3 4 6 1 2 3 6 4 12 5 20 30 1
𝟏 𝟐
𝟏 𝟑
𝟏 𝟔
𝟏 𝟒
𝟏 𝟏𝟐
𝟏 𝟓
𝟏 𝟐𝟎
𝟏 𝟑𝟎 1 1
Pascal
Leibniz 1 2 1 2 3 1 3 4 1 4 6 1 4 6 5
𝒇 𝒙 =𝟏 𝑥 4 + 𝟒 𝑥 3 + 𝟔 𝑥 2 +𝟒𝑥+𝟏
Factor GCF out of Leibniz Denominator
Form a polynomial of descending degree with Pascal’s coefficients and the derivative coefficients are the previous row in Leibniz’s triangle
𝑓 ′ 𝑥 =𝟒 𝑥 3 +𝟏𝟐 𝑥 2 +𝟏𝟐𝑥+𝟒

5. Formula To Find Any Term
Patterns
Formula To Find Any Term 1
𝟏 𝟐
𝟏 𝟑
𝟏 𝟔
𝟏 𝟒
𝟏 𝟏𝟐
𝟏 𝟓
𝟏 𝟐𝟎
𝟏 𝟑𝟎
𝟏 𝟔𝟎
R = Row
N = nth term 1
𝑹 𝑵 = 𝑹 −𝑵 ! 𝑵−𝟏 ! 𝑹!
𝟏 𝟐
𝟏 𝟐
𝟏 𝟏
Row 2
𝟒 𝟐 = 𝟒 −𝟐 ! 𝟐−𝟏 ! 𝟒! = 𝟐!𝟏! 𝟒! = 𝟏 𝟏𝟐
𝟏 𝟑
𝟏 𝟔
𝟏 𝟑
𝟏 𝟐
𝟐 𝟏
Row 3
𝟏 𝟒
𝟏 𝟏𝟐
𝟏 𝟏𝟐
𝟏 𝟒
𝟏 𝟑
𝟐 𝟐
𝟑 𝟏
Row 4
𝟓 𝟑 = 𝟓 −𝟑 ! 𝟑−𝟏 ! 𝟓! = 𝟐!𝟐! 𝟓! = 𝟏 𝟑𝟎
You can also get the terms by multiplying across. These fraction partitions show a factorial pattern.
𝟏 𝟓
𝟏 𝟐𝟎
𝟏 𝟑𝟎
𝟏 𝟐𝟎
𝟏 𝟓
𝟏 𝟒
𝟐 𝟑
𝟑 𝟐
𝟒 𝟏
Row 5
𝟏 𝟔
𝟏 𝟑𝟎
𝟏 𝟔𝟎
𝟏 𝟔𝟎
𝟏 𝟑𝟎
𝟏 𝟔
𝟏 𝟓
𝟐 𝟒
𝟑 𝟑
𝟒 𝟐
𝟓 𝟏
Row 6

6. Diagonals of Leibniz’s Triangle1
Diagonal 1
𝟏 𝟐
𝟏 𝟔
𝟏 𝟏𝟐
𝟏 𝟐𝟎
𝟏 𝟑𝟎
𝟏 𝟐
𝟏 𝟑
𝟏 𝟑
The Harmonic Series:
𝑛=1 ∞ 1 𝑛
Telescoping Series:
𝟏 𝟐 𝟏 𝟔 𝟏 𝟏𝟐 𝟏 𝟐𝟎 𝟏 𝟑𝟎
= 𝟏 𝟐 + 𝟏 𝟐 − 𝟏 𝟑 + 𝟏 𝟑 − 𝟏 𝟒 + 𝟏 𝟒 − 𝟏 𝟓 + 𝟏 𝟓 − 𝟏 𝟔
𝟏 𝟒
𝟏 𝟏𝟐
𝟏 𝟒
Telescoping Series:
𝟏 𝟒 𝟏 𝟐𝟎 𝟏 𝟔𝟎 𝟏 𝟏𝟒𝟎
= 𝟏 𝟒 + 𝟏 𝟏𝟐 − 𝟏 𝟑𝟎 + 𝟏 𝟑𝟎 − 𝟏 𝟔𝟎 + 𝟏 𝟔𝟎 − 𝟏 𝟏𝟎𝟓
Telescoping Series:
𝟏 𝟑 𝟏 𝟏𝟐 𝟏 𝟑𝟎 𝟏 𝟔𝟎 𝟏 𝟏𝟎𝟓
= 𝟏 𝟑 + 𝟏 𝟔 − 𝟏 𝟏𝟐 + 𝟏 𝟏𝟐 − 𝟏 𝟐𝟎 + 𝟏 𝟐𝟎 − 𝟏 𝟑𝟎 + 𝟏 𝟑𝟎 − 𝟏 𝟒𝟐
𝟏 𝟓
𝟏 𝟑𝟎
𝟏 𝟐𝟎 =𝟏
Diverges
𝟏 𝟔
𝟏 𝟔𝟎
𝟏 𝟔𝟎
= 𝟏 𝟐
= 𝟏 𝟑

7. Huygen’s challenge: Find the sum of the triangle reciprocals
Triangle Numbers: 1, 3, 6, 10,
Triangle Reciprocals: , , , ,
Leibniz Diagonal 2: , , , , Sum = 1
Factor out ½ : , , , ,
Triangle reciprocals
Since 𝟏 𝟐 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 𝒓𝒆𝒄𝒊𝒑𝒓𝒐𝒄𝒂𝒍𝒔 = 𝟏 ,
𝒕𝒉𝒆𝒏 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 𝒓𝒆𝒄𝒊𝒑𝒓𝒐𝒄𝒂𝒍𝒔 = 𝟐

8. Diagonal Series Patterns:
Diagonal Sum
𝟏 𝟐 2
+ 𝟏 𝟐
= 𝟏 𝟐 + 𝟏 𝟐 𝟏 𝟏
Triangle Reciprocals:
Tn = 2 𝑛(𝑛−1)
= 𝟏 𝟑 + 𝟏 𝟐 𝟏 𝟑 3
𝟏 𝟑
+ 𝟏 𝟔
𝟏 𝟒
+ 𝟏 𝟏𝟐
= 𝟏 𝟒 + 𝟏 𝟐 𝟏 𝟔 4
𝑺𝒖𝒎 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒊𝒐𝒓 𝒅𝒊𝒂𝒈𝒐𝒏𝒂𝒍𝒔:
Sn= 𝟏 𝒏 + 𝟏 𝟐 𝟐 𝒏(𝒏−𝟏)
𝑺𝒏= 𝟏 𝒏−𝟏

9. Thanks! ANY QUESTIONS? Summary: Exterior diagonals Harmonic Series
Interior diagonals Telescoping Series
Each Telescoping Sum Harmonic Series
Thanks!
ANY QUESTIONS?

10. THE END

11. References
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