<s>[INST] 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 $P.$
 [/INST] 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 $n \to \infty$?
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