[Guest]

Find the value of P in the following infinite product:

[Guest]

Find the value of P in the following infinite product:

2/3

All terms will cancel out except 3 in denominator from 1st term and 2 in numerator from 2nd term

[unreality]

Find the value of P in the following infinite product:

x^3 + 1 = (x+1)(x^2-x+1)

x^3 - 1 = (x-1)(x^2+x+1)

f(k) = (k - 1)*(kk + k + 1)

g(k) = (k + 1)*(kk - k + 1)

each term t(k) = f(k) / g(k)

we start with k=2, or 7/9 and proceed infinitely.

P = product(k = 2 to infinity) of t(k)

so we have for k=2:

f(k)/g(k) * f(k+1)/g(k+1) * f(k+2)/g(k+2) * f(k+3)/g(k+3) *...

expanding the first two fractions we get:

t(k) * t(k+1) = (k - 1)*(kk + k + 1) / (k + 1)*(kk - k + 1) * (k+1 - 1)*((k+1)^2 + k+1 + 1) / (k+1 + 1)*((k+1)^2 - k -1+ 1) * ....

= (k - 1)*(kk + k + 1) / (k + 1)(kk - k + 1) * (k)(kk + 2k + 1 + k + 1 + 1) / (k+2)(kk + 2k + 1 - k - 1 + 1)

= (k - 1)*(kk + k + 1) / (k + 1)(kk - k + 1) * (k)(kk + 3k + 3) / (k+2)(kk + k + 1)

= (k-1)(k) (kk + k + 1) (kk + 3k + 3) / [ (k+1)(kk - k + 1)(k+2)(kk + k + 1) ]

= (k-1)(k) / (k+1)(k+2) * (kk + 3k + 3) / (kk - k + 1)

= 1/6 (4 +6 + 3)/(4-2+1)

= 13/18 in the case of 2, which matches so far t(2) * t(3)

so let's say h(k) = t(k) * t(k+1)

now jump forward by two k's. The new h(k) has the same opening thing: (k-1)(k) / (k+1)(k+2)

except here, the (k-1) and (k) is equal to the (k+1) and (k+2) of the previous h(k), and therefore cancels out, leaving only the initial (k-1)(k) on that far side which is (1)(2) = 2.

As for the second half of the h(k) formula, let's apply h(k+2)

(kk + 4k + 4 + 3k + 6 + 3) / (kk + 4k + 4 - k - 2 + 1) = (kk + 7k + 13) / (kk + 3k + 3)

thus the (kk + 3k + 3) cancels out, etc. Each new term cancelling the one before. Leaving, for this half of the equation, only the original divison by (kk - k + 1) which is (4 - 2 + 1) = 3

so we're left with (2)/(3) = 2/3

there's probably a way easier way to do this haha

[Guest]

its 2/3 simple maths...

Find the value of P in the following infinite product:

[Guest]

Find the value of P in the following infinite product:

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.