Compute the infinite series of the $\displaystyle{\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n}.}$

Question

Can somebody help me with this.

Compute the infinite series of the $\displaystyle{\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n}.}$

I have no idea how to start with this. This is my first time encounter with the problem. I'm sorry. If you could, I will appreciate this.

What I did by the way is to expand the product. However, is there a way to do this in a way that it does not like doing the expanding classical form?

Answer 1

Your product is $\prod_n\frac{(n+1)/(n+2)}{n/(n+1)}$, which telescopes to $2$.

Answer 2

Is this acceptable or can somebody point out my mistake?

$\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n} = \lim_{n \rightarrow \infty} \prod_{k=1}^n \frac{(k+1)^2}{k^2+2k} = \lim_{n\rightarrow \infty} \left[\frac{2{(2)}}{1{(3)}} \cdot \frac{{(3)}(3)}{{(2)}(4)} \cdots \frac{(n-1){(n-1)}}{(n-2){n}} \cdot \frac{{(n)}{(n)}}{{(n-1)}{(n+1)}} \cdot \frac{(n+1)(n+1)}{n(n+2)} \right] = \lim_{n \rightarrow \infty} \frac{2(n+1)}{n+2} = \lim_{n\rightarrow \infty} \frac{2\left(1 + 1/n\right)}{1 + 2/n} = 2$.