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I am looking for some advice about approaching the following computations:

$$\lim\limits_{n\to\infty}{\prod\limits_{k=1}^{n}{\Bigr(1-\frac{2}{(k+1)\cdot(k+2)}\Bigr)}}=\large{?}$$ $$\lim\limits_{n\to\infty}{\prod\limits_{k=1}^{n}{\Bigr(1+\frac{2}{(k+1)\cdot(k+2)}\Bigr)}}=\large{?}$$

I tried to look for the logarithm of the limit and Taylor series, but nothing good came of it.
Thank you in advance.

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    $\begingroup$ What have tried? C'mon, are you gonna throw a problem at us just like that? $\endgroup$
    – user369582
    Sep 29, 2016 at 20:04
  • $\begingroup$ No, I tried to look for the logarithm of the limit and Taylor series. But nothing good came of it. I need an idea or tip. $\endgroup$
    – Roman
    Sep 29, 2016 at 20:08
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    $\begingroup$ People should go easier on downvoting questions from new users. Suggestions about improving questions are fine, but you should leave the OP at least 10 seconds for editing his question, don't you? $\endgroup$ Sep 29, 2016 at 20:14

1 Answer 1

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The trick for tackling the first limit is to recognize a telescopic product in disguise. Since $$ 1-\frac{2}{(k+1)(k+2)}=\frac{k(k+3)}{(k+1)(k+2)}\tag{1} $$ we have $$ \prod_{k=1}^{n}\left(1-\frac{2}{(k+1)(k+2)}\right) = \prod_{k=1}^{n}\frac{k}{k+1}\prod_{k=1}^{n}\frac{k+3}{k+2}=\frac{1}{n+1}\cdot\frac{n+3}{3} \tag{2}$$ hence the first limit equals $\color{red}{\large{\frac{1}{3}}}$. The second limit is trickier. You may exploit Wallis' product and the Weierstrass product for the $\cosh$ function, $$ \cosh(z)=\prod_{n\geq 0}\left(1+\frac{4z^2}{(2n+1)^2 \pi^2}\right)\tag{3} $$ evaluated at a peculiar $z$, to deduce that: $$ \prod_{k\geq 1}\left(1+\frac{2}{(k+1)(k+2)}\right) =\color{red}{\frac{1}{4\pi}\cosh\left(\frac{\pi\sqrt{7}}{2}\right)}.\tag{4}$$

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  • $\begingroup$ Thank you! I did not see the telescope product. Everything is very simple. PS Sorry for the error when creating a question $\endgroup$
    – Roman
    Sep 29, 2016 at 20:22
  • $\begingroup$ @RomanLevin: you are welcome, and there is no need to apologize. We are here to learn, each one of us. $\endgroup$ Sep 29, 2016 at 20:23

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