Find the limit of the following product series

Question

I am trying to find the limit of the following :

$$\lim_{n\to \infty} \left(1- \frac{ 2}{3}\right)^\frac{3}{n} \left(1- \frac{ 2}{4}\right)^\frac{4}{n} \left(1- \frac{ 2}{5}\right)^\frac{5}{n}...\left(1- \frac{ 2}{n+2}\right)^\frac{n+2}{n} $$

Answer 1

$a_n =(1- \frac{ 2}{3})^\frac{3}{n}(1- \frac{ 2}{4})^\frac{4}{n}(1- \frac{ 2}{5})^\frac{5}{n}...(1- \frac{ 2}{n+2})^\frac{n+2}{n} =\prod_{k=3}^{n+2} (1-\frac{2}{k})^{k/n} $.

$\begin{array}\\ b_n &= \ln a_n\\ &= \sum_{k=3}^{n+2} \frac{k}{n}\ln(1-\frac{2}{k})\\ &= \frac1{n}\sum_{k=3}^{n+2} k\ln(\frac{k-2}{k})\\ &= \frac1{n}\sum_{k=3}^{n+2} k(\ln(k-2)-\ln(k))\\ &= \frac1{n}\left(\sum_{k=3}^{n+2} k\ln(k-2)-\sum_{k=3}^{n+2} k\ln(k)\right)\\ &= \frac1{n}\left(\sum_{k=1}^{n} (k+2)\ln(k)-\sum_{k=3}^{n+2} k\ln(k)\right)\\ &= \frac1{n}\left(\sum_{k=1}^{n} k\ln(k)+\sum_{k=1}^{n} 2\ln(k)-\sum_{k=3}^{n+2} k\ln(k)\right)\\ &= \frac1{n}\left(1\ln(1)+2\ln(2)+2\sum_{k=1}^{n} \ln(k)-(n+1)\ln(n+1)-(n+2)\ln(n+2)\right)\\ &= \frac1{n}\left(2\ln(2)+2\ln(n!)-(n+1)\ln(n+1)-(n+2)\ln(n+2)\right)\\ &= \frac1{n}\left(2\ln(2)+2(n\ln(n)-n+O(\ln(n))-n(\ln(n+1)+\ln(n+2))-(\ln(n+1)+2\ln(n+2))\right)\\ &= 2\ln(n)-(\ln(n+1)+\ln(n+2))-2+O(\ln(n)/n)-(\ln(n+1)+2\ln(n+2))\\ &= \ln(n^2/((n+1)(n+2))-2+O(\ln(n)/n)\\ &= -\ln((n+1)(n+2)/n^2)-2+O(\ln(n)/n)\\ &= -\ln((n^2+3n+2)/n^2)-2+O(\ln(n)/n)\\ &= -\ln(1+3/n+2/n^2)-2+O(\ln(n)/n)\\ &= -(3/n+O(1/n^2))-2+O(\ln(n)/n)\\ &= -2+O(\ln(n)/n)\\ &\to -2\\ \end{array} $

so $a_n \to e^{-2} $.

$P(error) > .27$.

Answer 2

You can rewrite it as

$$a_n = \sqrt[n]{\prod_{j=1}^n\left(1-\frac{2}{j+2} \right)^{j+2}} $$

As this is of the form $\sqrt[n]{\prod_{j=1}^n c_j}$, you have

$$ \lim_n a_n = \lim_j c_j = e^{-2} $$

Answer 3

By setting $a_n = \log\left(1-\frac{2}{n}\right)^n$ for any $n\geq 3$ we have $\lim_{n\to +\infty} a_n=-2.$
By Cesàro theorem (convergence implies convergence in average to the same limit) we have $$ \lim_{n\to +\infty}\frac{a_3+a_4+\ldots+a_{n+2}}{n} = -2$$ and by exponentiating both sides $$ \lim_{n\to +\infty}\left(1-\frac{2}{3}\right)^{\frac{3}{n}}\cdots\left(1-\frac{2}{n+2}\right)^{\frac{n+2}{n}} =\color{red}{e^{-2}}.$$