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 Infinite Product (Posted on 2013-06-24)
Find the value of the infinite product
P=7/9*26/28*63/65*....*(k3-1)/(k3+1)*...

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Analytic solution Comment 3 of 3 |
consider the kth product

(k³ - 1) / (k³ + 1)

factor the numerator:

(k-1)(k²+k+1)

consider the k+1th product:

((k+1)³ - 1) / ((k+1)³ + 1)

expand the denominator:

k³ + 3k² + 3k + 2

and factor: (k+2)(k² + k + 1)

So the quadratic term in the kth numerator cancels the quadratic term in the k+1th demoninator, leaving a product (k-1)*k*(k+1)... in the numerator, and a product (k+2)(k+3)... in the denominator, AND the very first denominator which has no prior numerator to cancel.

All of the (k+n) terms in the denominator cancel corresponding terms in the numerator, leaving just three terms in the numerator without matches: (k-1)*k*(k+1), and the very first denominator is still left. That means the product overall is equal to:

(K³ - K) / (K³ + 1) where K is the very first value of k: 2 for this problem.

Substituting, the product is (8-2) / (9) = 6/9 = 2/3

 Posted by Paul on 2013-06-24 18:07:53

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