Finding the finite product $\prod_{k=2}^{\infty}\frac{k^2}{k^2-1}$. Let $\displaystyle\prod_{k=2}^{\infty}\frac{k^2}{k^2-1}=P^2$. I need hint to find $P.$
 A: You have
$$P^2=\lim_{n\to\infty}\prod_{k=2}^n\frac{k^2}{k^2-1}=\lim_{n\to\infty}\left(n!\right)^2\prod_{k=2}^n\frac1{(k+1)(k-1)}=\lim_{n\to\infty}\left(n!\right)^2\frac{2n}{n!n!(n+1)}=\lim_{n\to\infty}\frac{2n}{n+1}=2$$
Thus $P=\pm\sqrt2$.
A: We can split this up into two fractions like this:
$$\prod_{k=2}^\infty \frac{k}{k-1}\cdot\frac{k}{k+1}$$
Now, if we go to $k+1$, the next term is:
$$\frac{k+1}{k} \cdot \frac{k+1}{k+2}$$
Now, notice how if we multiply these two terms, the two middle fractions cancel:
$$\left(\frac{k}{k-1}\cdot \frac{k}{k+1}\right)\cdot \left(\frac{k+1}{k}\cdot \frac{k+1}{k+2}\right)=\frac{k}{k-1}\cdot \frac{k+1}{k+2}$$
Now, if you multiply on $k+2$, the middle fractions cancels again:
$$\left(\frac{k}{k-1}\cdot \frac{k+1}{k+2}\right)\cdot\left(\frac{k+2}{k+1}\cdot \frac{k+2}{k+3}\right)=\frac{k}{k-1}\cdot \frac{k+2}{k+3}$$
Thus, the partial product is simply the first fraction, which is $\frac{2}{2-1}=2$ times whatever the last fraction is. Now, the last fraction is always $\frac{n}{n+1}$ where $n$ is the top number in the partial product. Thus, the formula for the partial product is:
$$\frac{2n}{n+1}$$
Now, what happens to this as $n \to \infty$?
A: Hint:
$$ \frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1}$$
is the Weierstrass product for the sine function. We are dealing with:
$$ \prod_{k\geq 2}\left(1-\frac{1}{k^2}\right)^{-1} = \lim_{z\to 1^-}\frac{\pi z (1-z^2)}{\sin(\pi z)}=\color{red}{2}.\tag{2} $$
A: Hint: Take log. It becomes  (partial sum)
$$\sum_{k=2}^n 2\log(k) - \log(k-1)-\log(k+1)$$
Do you see two telescoping sums?
