Infinite Product

ThudanBlunder
Posers and Puzzles 23 Aug '06 03:57
1. 23 Aug '06 03:57
Find the value of the infinite product (7/9)*(26/28)*(63/65)*........*(n^3 - 1)/(n^3 + 1) * .........
2. PBE6
Bananarama
23 Aug '06 15:49
Originally posted by ThudanBlunder
Find the value of the infinite product (7/9)*(26/28)*(63/65)*........*(n^3 - 1)/(n^3 + 1) * .........
We can factor both the top and bottom expressions as follows:

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

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

We'll deal with each part of the factorization seperately, starting with the linear part. As we move through the indices, the linear factor in the numerator becomes (n-1), n, (n+1), (n+2), etc... while the linear factor in the denominator becomes (n+1), (n+2), etc... As you can see, once the linear factor in the numerator becomes (n+1), they start cancelling out. The leftovers in the numerator are:

(n-1)(n)

Now for the quadratic part. As we move through the indices, the quadratic in the denominator becomes (n^2-n+1), ((n+1)^2-(n+1)+1) = (n^2+2n+1-n-1+1) = (n^2+n+1), etc... which cancels out the quadratic in the numerator (n^2+n+1), ((n+1)^2+(n+1)+1), etc... The leftovers in the deonominator are:

(n^2-n+1)

So we end up with:

P = (n-1)(n)/(n^2-n+1)

This product starts with n=2 as given in the question where the first term is (2^3-1)/(2^3+1) = (7/9), so we evaluate the above expression with n=2 to get:

P = (2-1)(2)/(2^2-2+1) = (2/3)
3. 23 Aug '06 17:474 edits
Originally posted by ThudanBlunder
Find the value of the infinite product (7/9)*(26/28)*(63/65)*........*(n^3 - 1)/(n^3 + 1) * .........
I agree. I had, after completing the square in n^2 + n + 1 and n^2 - n + 1:

( n^3 - 1 ) / ( n^3 + 1 ) = ( ( n - 1) / ( n + 1 ) ) * ( ( ( n + 1/2 )^2 + 3/4 ) / ( ( n - 1/2 )^2 + 3/4 ) )

so the product of the first N terms telescopes to

( 2 / ( N * ( N + 1 ) ) ) * ( ( ( N + 1/2 )^2 + 3/4 ) / ( ( 1 + 1/2 )^2 + 3/4 ) ).

This tends to 2/3 as N tends to infinity.