Limit of $\prod_{k=1}^n \frac{(k^2)}{(k+1)^2}$

Question

$$\prod_{k=1}^\infty \frac{(k^2)}{(k+1)^2}$$

What is this product? Because of intuition I would say that this is 0, but if you expand this, you can cancel the fractions, so I was wondering, what happens at infinity?

Answer 1

Hint:

$$\prod_{i=1}^n\dfrac{f(i)}{f(i+1)}=\dfrac{f(1)}{f(n+1)}$$

here $f(x)=x^2$

Now set $n\to\infty$

Answer 2

HINT

Note that

$$\prod_{i=1}^n \frac{(i^2)}{(i+1)^2}=\frac 1 {2^2}\cdot \frac {2^2} {3^2}\cdot \frac {3^2} {4^2}\cdot...\cdot \frac {(n-1)^2} {n^2}\cdot \frac {n^2} {n+1^2}=\frac 1 {(n+1)^2}$$