Compute: $\lim_{n \to \infty } \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}$ Help me please to compute the limit of:
$
\lim_{n \to \infty }  \left ( 1-\frac{2}{3} \right ) ^{\frac{3}{n}}\cdot  \left ( 1-\frac{2}{4} \right ) ^{\frac{4}{n}}\cdot  \left ( 1-\frac{2}{5} \right ) ^{\frac{5}{n}}\cdots \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}} 
$
What I did:
$
L=\frac{a_{n+1}}{a_n} =   \frac{ \left ( 1-\frac{2}{n+3} \right ) ^{\frac{n+3}{n+1}}}{ \left ( 1-\frac{2}{n+2} \right ) ^{\frac{n+2}{n}}}=\frac{2/e}{2/e}=1
$
But it's not. Since $L=1$, I need use something else...
 A: $(1-2/k)^k$ approaches $e^{-2}$, so most of the factors are near $e^{-2}$.
Then you take the $n$th root of them all, and there are $n$ factors, so the answer is $e^{-2}$
A: Let
$$f_n = \left ( 1-\dfrac{2}{3} \right ) ^{\dfrac{3}{n}}\cdot  \left ( 1-\dfrac{2}{4} \right ) ^{\dfrac{4}{n}}\cdot  \left ( 1-\dfrac{2}{5} \right ) ^{\dfrac{5}{n}}\cdot \cdot \cdot   \left ( 1-\dfrac{2}{n+2} \right ) ^{\dfrac{n+2}{n}} $$
We have
$$\ln(f_n) = \sum_{k=1}^n \dfrac{k+2}{n}\ln\left(1-\dfrac2{k+2}\right)$$
Hence,
\begin{align}
n \ln(f_n) & = \sum_{k=1}^n (k+2)\left(\ln(k)-\ln(k+2)\right)=\sum_{k=1}^n (k+2)\ln(k) - \sum_{k=1}^n(k+2)\ln(k+2)\\
& = \sum_{k=1}^n (k+2)\ln(k) - \sum_{k=3}^{n+2}k\ln(k)\\
& = 3\ln(1) + 4\ln(2) + \sum_{k=3}^n (k+2) \ln(k) - \sum_{k=3}^n k \ln(k) - (n+1)\ln(n+1) - (n+2)\ln(n+2)\\
& = 2 \ln(2) + 2 \sum_{k=2}^n \ln(k) - (n+1)\ln(n+1) - (n+2)\ln(n+2)\\
& = 2 \ln(2) + 2 \ln(n!) - (n+1)\ln(n+1) - (n+2)\ln(n+2)\\
& = 2 \ln(2) + \ln(2\pi n) + 2n \ln(n) - 2n - (n+1)\ln(n+1) - (n+2)\ln(n+2) + \mathcal{O}(1/n)\\
\ln(f_n) & = -2 + \mathcal{O}(\ln(n)/n)
\end{align}
Hence, $f_n \to e^{-2}$.
