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6. Show $\alpha _n := \prod_{j=1}^n 1-\frac{1}{3^j}$ converges, with it limit positive. 

[Solution] It' trivial that $(\alpha_n)$ decreases, with $0$ its lower bound. It converges. I'll give two ways to show that it has a positive lower bound. 

$\text{ }$

(Method I) We know that $(1-a)(1-b)\textgreater 1-(a+b)$ for $a,b\in (0,1)$. This means 
\begin{align}
&(1-\frac{1}{3^1})(1-\frac{1}{3^2})(1-\frac{1}{3^3})...(1-\frac{1}{3^n})\notag\\
&\textgreater [1-(\frac{1}{3^1}+\frac{1}{3^2})] (1-\frac{1}{3^3})...(1-\frac{1}{3^n})\notag\\
&\textgreater [1-(\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3})]...(1-\frac{1}{3^n})\notag\\
&\textgreater ... \textgreater 1- (\frac{1}{3^1}+\frac{1}{3^2}+...+ \frac{1}{3^n})\notag\\
&= 1- \frac{1}{3}\frac{1- (\frac{1}{3})^n}{1-\frac{1}{3}}\textgreater \frac{1}{2}\notag
\end{align}
So we conclude that the limit might greater than $0$. 

$\text{ }$

(Method II) By an inequality $3^n\textgreater n^2$, we find that $1-\frac{1}{3^j}\textgreater 1-\frac{1}{j^2}$. Hence
\begin{align}
&\prod_{j=1}^n 1-\frac{1}{3^j} \textgreater \frac{2}{3} \prod_{j=2}^n 1-\frac{1}{j^2} = \prod_{j=1}^n (1-\frac{1}{j})(1+\frac{1}{j})\notag\\
&= \frac{2}{3} (\frac{1}{2} \frac{2}{3} ... \frac{n-1}{n})(\frac{3}{2}\frac{4}{3} ... \frac{n+1}{n})\notag\\
&= \frac{n+1}{3n}\textgreater \frac{n}{3n} =\frac{1}{3}\notag
\end{align}
Similarly we finish the proof. 


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