2015/116) Find the limit of the product as k goes to infinite $\prod_{n=1}^{k}\frac{n^2}{n^2-1}$

We propose $\prod_{n=1}^{k}\frac{n^2}{n^2-1}=\frac{2k}{k+1}$

for n =2 we have product = $\frac{4}{3} = \frac{2 * 2}{2+1}$

let it be true of n = k

so we have $\prod_{n=1}^{k}\frac{n^2}{n^2-1}=\frac{2k}{k+1}$

multiply by (k+1)st term to get   $f(k+1) = \prod_{n=1}^{k}\frac{n^2}{n^2-1} * \frac{(k+1)^2}{(k+1)^2 - 1}$
= $\frac{2k}{k+1} * \frac{(k+1)^2}{k(k+2)} = \frac{2 * (k +1)}{(k+2)}$
so if it is true for k it is true for k + 1
Now that we have found the closed form as k goes to infinite above product goes to 2 (converges to 2)