Can you explain what is going on here? Where might Leibniz want to be a bit more rigorous?
In this activity we investigate some of the series that Leibniz investigated.
Series pop-up at an early age. I distinctly remember being in fourth grade, sitting at my desk, starring at my ruler, wondering how \(1/3\) of a foot could simultaneously be \(4\) inches (clearly a finite number) and \(0.333333\dots \) of a foot (a number that somehow seemed finite and infinite at the same time). I was struggling with the implicit concept that
\[ \frac {1}{3} = \frac {3}{10}+\frac {3}{100}+\frac {3}{1000}+\frac {3}{10000}+\cdots \]
Leibniz (and other mathematicians of the era) had similar feelings regarding series. Leibniz’s mentor, Christian Huygens, suggested that Leibniz work on computing the sum of the reciprocal of the triangular numbers. Recall that the triangular numbers are the number of dots in discrete equilateral triangles:
![[Picture]](leibnizAndSeries-5ab5466127c173a69b29bc93c745c42f.svg)
Consider Leibniz’s “proof.” \begin{align*} S &= 1+ \frac {1}{3} + \frac {1}{6} + \frac {1}{10} + \frac {1}{15} + \frac {1}{21} + \cdots \\ \frac {S}{2}&=\frac {1}{2} + \frac {1}{6} + \frac {1}{12} + \frac {1}{20} + \frac {1}{30}+\frac {1}{42} + \cdots \\ \frac {S}{2}&=\left (1-\frac {1}{2}\right ) + \left (\frac {1}{2}-\frac {1}{3}\right ) + \left (\frac {1}{3}-\frac {1}{4}\right ) + \left (\frac {1}{4}-\frac {1}{5}\right ) + \cdots \\ \frac {S}{2}&=1+\left (-\frac {1}{2} + \frac {1}{2}\right )+\left (-\frac {1}{3} + \frac {1}{3}\right )+\left (-\frac {1}{4} + \frac {1}{4}\right )+ \cdots \\ \frac {S}{2}&= 1\\ S &= 2. \end{align*}
Now consider this summation of
\[ 1+ \frac {1}{3} + \frac {1}{6} + \frac {1}{10} + \frac {1}{15} + \frac {1}{21} + \cdots \]
Write \begin{align*} \sum _{k=1}^\infty \frac {2}{k(k+1)} &= 1+ \frac {1}{3} + \frac {1}{6} + \frac {1}{10} + \frac {1}{15} + \frac {1}{21} + \cdots \\ \sum _{k=1}^n \frac {2}{k(k+1)} &= \sum _{k=1}^n 2\left (\frac {1}{n}- \frac {1}{n+1}\right )\\ \sum _{k=1}^n \frac {2}{k(k+1)} &= 2\left (1 - \frac {1}{n+1}\right )\\ \lim _{n\to \infty } \sum _{k=1}^n \frac {2}{k(k+1)} &= \lim _{n\to \infty }2\left (1 - \frac {1}{n+1}\right )\\ \sum _{k=1}^\infty \frac {2}{k(k+1)} &= 2. \end{align*}
What’s going on here? How does this compare to Leibniz’s proof?
In addition to (co)inventing calculus, Leibniz, dreamed up this beast while trying to solve problems like the one posed to him by Huygens:
\[ \begin{array}{@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}c@{\,}} & & & & \frac {1}{1} & & & & \\ & & & \frac {1}{2} & & \frac {1}{2} & & & \\ & & \frac {1}{3} & & \frac {1}{6} & & \frac {1}{3} & & \\ & \frac {1}{4} & & \frac {1}{12} & & \frac {1}{12} & & \frac {1}{4} &\\ \frac {1}{5} & & \frac {1}{20} & & \frac {1}{30} & & \frac {1}{20} & & \frac {1}{5} \end{array} \]
Explain how the following expression
\[ \frac {1}{r\cdot \binom {r-1}{c-1}} \]
corresponds to entries of the harmonic triangle. Feel free to draw diagrams
and give examples. Explain how the harmonic triangle is formed. In your explanation, use the notation
\[ \frac {1}{r\cdot \binom {r-1}{c-1}} \]
If you were so
inclined to do so, could you state a single equation that basically encapsulates your explanation above?
Explain how to use the harmonic triangle to go from:
\[ \frac {1}{2} + \frac {1}{6} + \frac {1}{12} + \frac {1}{20}+\cdots \]
to \[ \left (1 - \frac {1}{2}\right ) + \left (\frac {1}{2}-\frac {1}{3}\right ) + \left (\frac {1}{3}- \frac {1}{4}\right ) + \left (\frac {1}{4}-\frac {1}{5}\right ) + \cdots \]
Conclude by explaining why Leibniz said: \[ \frac {1}{2} + \frac {1}{6} + \frac {1}{12} + \frac {1}{20}+\cdots = 1 \]
Can you generalize the results above? Can you give a list of infinite sums and conjecture what they will converge
to?
2025-01-06 15:59:30