<s>[INST] Find $\lim_{n \to \infty} \prod_{k=1}^{n} \frac{(k+1)^2}{k(k+2)}$ I have to find the following limit:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}}$$
This is what I tried:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = 
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{k^2+2k+1}{k^2+2k}} = 
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \bigg(\dfrac{k^2+2k}{k^2+2k}} + \dfrac{1}{k^2+2k} \bigg ) = $$
$$ = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \bigg(1 } + \dfrac{1}{k^2+2k} \bigg ) = 
\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 }  + \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} }$$
Now, 
$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 } = 1$
and:
$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} } = 0$
I think the above equals $0$, since this is a product and the limit of the last term of the product is $0$, so the whole thing would be $0$, but I am not exactly sure if my intuition is right.
So that means:
$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} 1 }  + \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{1}{k^2+2k} } = 1 + 0 = 1$$
The problem I have is that my textbook claims that the correct answer is $2$, not $1$. So I did something wrong, however, I can't spot my mistake/mistakes.
 [/INST] First observe that the product is a telescopic product: $$\prod_{k = 1}^n \frac{(k+1)^2}{k(k+2)} = \frac{2^2}{3} \cdot \frac{3^2}{2\cdot 4} 
\cdot \frac{4^2}{3 \cdot 5} \cdot ... \cdot \frac{n^2}{(n-1)(n+1)} \cdot \frac{(n+1)^2}{n(n+2)} = \frac{2(n+1)}{n+2}. $$
In case of need use induction to prove it.
Now it is easy:
$$ \lim_{n \to \infty} \frac{2(n+1)}{n+2} = 2.$$
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