<s>[INST] How do I calculate the limit for this multiplication? $$\lim_{n\to\infty}\left(1-\frac{2}{3}\right)^{\tfrac{3}{n}}\cdot\left(1-\frac{2}{4}\right)^{\tfrac{4}{n}}\cdot\left(1-\frac{2}{5}\right)^{\tfrac{5}{n}}\cdots\left(1-\frac{2}{n+2}\right)^{\tfrac{n+2}{n}}$$
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I mean,I tried to use Sandwitch rule but it didnt work.
 [/INST] The $m^{th}$ term is
\begin{align}
a_m & = \prod_{n=1}^m \left(1-\dfrac2{n+2}\right)^{(n+2)/m} = \prod_{n=1}^m \left(\dfrac{n}{n+2}\right)^{(n+2)/m} = \left(\dfrac{\prod_{n=1}^m n^{(n+2)/m}}{\prod_{n=1}^m (n+2)}\right)^{(n+2)/m}\\
& = \left(\dfrac{\prod_{n=1}^m n^{n/m}}{\prod_{n=3}^{m+2} n^{n/m}}\right) \prod_{n=1}^m n^{2/m} = \dfrac{2^{2/m}(m!)^{2/m}}{(m+1)^{(m+1)/m}(m+2)^{(m+2)/m}}
\end{align}
This gives us that
\begin{align}
\log(a_m) & = \dfrac{2\log2}m + \dfrac{2\log(m!)}m - \dfrac{m+1}m \log(m+1) - \dfrac{m+2}{m+1} \log(m+2)\\
& = \dfrac{2\log2}m + \dfrac{2}m\left(m\log(m)-m + \mathcal{O}(\log(m))\right) - \dfrac{m+1}m \log(m+1) - \dfrac{m+2}{m+1} \log(m+2)\\
\end{align}
where the last step relies on DeMoivre's identity.
Finishing step $1$:

 This gives us$$\log(a_m) = -2 + 2 \log(m) - \log(m+1)-\log(m+2) + \mathcal{O}\left(\dfrac{\log(m)}m \right)$$

Finishing step $2$:

 Taking $m \to \infty$, we obtain$$\lim_{m \to \infty} \log(a_m) = -2$$

</s>