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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.

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  • $\begingroup$ You have something like $$\frac{k+1}{k}\cdot \frac{k+1}{k+2}\cdot \frac{k+2}{k+1}\cdot \frac{k+2}{k+3}=\frac{k+1}{k}\cdot \frac{k+2}{k+3}$$ which should help with simplifying the expression. $\endgroup$
    – rtybase
    Nov 20 '19 at 20:46
  • $\begingroup$ The problem in your attempt is when you break the product into a sum of two products. This is probably based on an intuition from working with sums. With sums, splitting things like that makes sense because addition is commutative with itself, for example (1+2)+(4+3) could just as easily be written (1+4)+(2+3). However in products this splitting really doesn’t make sense, it’s like if I tried to claim (1+4)•(2+3) = (1•2)+(4•3), which is clearly not true now that we think of it in more familiar terms. $\endgroup$
    – Robo300
    Nov 20 '19 at 21:23
  • $\begingroup$ The way you multiplied the terms in the third line is not correct because $(1+x)(1+y)(1+z) \neq 1 \times 1 \times 1 + xyz$ $\endgroup$
    – Alex
    Nov 20 '19 at 23:05
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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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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}

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

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