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.
 A: Hint: Since $1+x\le e^x$, we have that $1-\frac{x}{1+x}\le e^{-\frac{x}{1+x}}\implies e^{\frac{x}{1+x}}\le1+x$. Therefore,
$$
e^{-2/k}\le1-\frac2{k+2}\le e^{-2/(k+2)}
$$
and, thus,
$$
e^{-2\left(1+\frac2k\right)}\le\left(1-\frac2{k+2}\right)^{k+2}\le e^{-2}
$$
Use this to bound the product. Then take $n^{\text{th}}$ roots and apply the Squeeze Theorem.
A: Hint: Rewrite your quantity as 
$$
\exp\left(\sum_{k=3}^{n+2}\frac{k}{n} \ln\left(1-\frac{k-1}{k}\right) \right)
= \exp\left(\sum_{k=3}^{n+2}\frac{k}{n} \ln\left(1-\frac{2}{k}\right) \right)
$$
and use Cesàro's theorem to find the limit of $\sum_{k=3}^{n+2}\frac{k}{n} \ln\left(1-\frac{2}{k}\right)$ before concluding by continuity of the exponential.

 For Cesàro's theorem, observe that $a_k = k\ln(1-\frac{2}{k})\xrightarrow[k\to\infty]{} -2$.

A: 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$$\lim_{m \to \infty} \log(a_m) = -2$$

