Find $\lim_{n \to \infty} \prod_{k=1}^{n} \frac{(k+1)^2}{k(k+2)}$

Question

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.

Answer 1

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: $$ \lim_{n \to \infty} \frac{2(n+1)}{n+2} = 2.$$

Answer 2

Using $$\prod_{k=1}^{n} (k+1) = \prod_{k=2}^{n+1} k = (n+1)!$$ then, with similar products, \begin{align} L &= \lim_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} \\ &= \lim_{n \to \infty} \frac{2 \, ((n+1)!)^2}{n! \, (n+2)!} \\ &= \lim_{n \to \infty} \frac{2 (n+1)}{n+2} = 2 \end{align}

Answer 3

$$\lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{(k+1)^2}{k(k+2)}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \dfrac{k+1}{k}\dfrac{k+1}{k+2}} = \lim\limits_{n \to \infty} {\displaystyle \prod_{k=1}^{n} \frac{ \dfrac{k+1}{k}}{\dfrac{k+2}{k+1}}}$$ $$ = \lim\limits_{n \to \infty} \left( \frac{\frac{1+1}{1}}{\frac{1+2}{1+1}} \frac{\frac{2+1}{2}}{\frac{2+2}{2+1}}\frac{\frac{3+1}{3}}{\frac{3+2}{3+1}}\cdots \frac{\frac{n+1}{n}}{\frac{n+2}{n+1}} \right) = \lim\limits_{n \to \infty} \left( \frac{\frac{1+1}{1}}{\frac{n+2}{n+1}} \right)$$ $$ = \lim\limits_{n \to \infty} \left( 2 \frac{n+1}{n+2} \right) = 2 $$