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Solution to puzzle 15: Infinite product

Find the value of the infinite product

P = 7/9 * 26/28 * 63/65 * ... * (k^3 - 1)/(k^3 + 1) * ...

Factorizing numerator and denominator, we have

k3 − 1 = (k − 1)(k2 + k + 1)
k3 + 1 = (k + 1)(k2 − k + 1)

Note that k2 − k + 1 = (k − 1)2 + (k − 1) + 1, and so k3 + 1 = [(k − 1) + 2][(k − 1)2 + (k − 1) + 1], allowing cancellation of the quadratic factor across successive terms, and of the linear factor across "next but one" terms.

We can now calculate Pn, the partial product of the first n − 1 terms.

P(n) = 7/9 * 26/28 * 63/65 * ... * (n^3 - 1)/(n^3 + 1) = (1/3 * 7/3) * (2/4 * 13/7) * (3/5 * 21/13) * ... * [(n-1)/(n+1) * (n^2+n+1)/(n^2-n+1)] = (2/3) * [(n^2+n+1)/(n(n+1))] = (2/3) * [1 + 1/(n(n+1))]

As n tends to infinity, Pn tends to 2/3.
That is, the infinite product, P, converges to 2/3; P = Pinfinity = 2/3.


Remarks

Letting w = −1/2 + iroot 3/2 be a complex cube root of unity, we have

k3 − 1 = (k − 1)(k − w)(k + w + 1)
k3 + 1 = (k + 1)(k + w)(k − w − 1)

This shows explicitly that k2 − k + 1 = (k − 1)2 + (k − 1) + 1, and how to telescope the partial product.


Further reading

  1. A Collection of Infinite Products – I
  2. Pentagonal Number Theorem

Source: Infinite product, equation 10

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