2011/089) Evaluate the infinite product Π(n=2..∞)n^2/(n^2-1).

First we try to get a closed form for the product

We have t(n) = n^2/((n+1)(n-1))

Π(n=2) n^2/(n^2-1)= 2^2/(3)

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

Π(n=2..4)n^2/(n^2-1)= 2*3/(4) * 4^2 /(3*5) = 2 * 4 / 5

From the pattern we feel that

Π(n=2..k)n^2/(n^2-1) = 2 * k/(k+1)

It is true for n = 2

If it is true for n =k then

Π(n=2..k+1)n^2/(n^2-1) = 2 * k/(k+1) * (k+1)^2/(k*(k+2) = 2 *(k+1)/(k+2)

So it is true for k+ 1

Π(n=2..k)n^2/(n^2-1) = 2k/(k+1)

Now that we have found the closed form as k goes to inifinite above product goes to 2 (converges to 2)