find $\lim_{n\to\infty}(1+\frac{1}{3})(1+\frac{1}{3^2})(1+\frac{1}{3^4})\cdots(1+\frac{1}{3^{2^n}})$ 
$$\lim_{n\to\infty}\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^4}\right)\cdots\left(1+\frac{1}{3^{2^n}}\right)$$

You have to find the given limit when $n$ tends to infinity. I can't think of a way I tried taking $\log$ and proceeding but nothing helped please help. 
 A: HINT:
$$(1-x)(1+x)=1-x^2$$
$$(1-x^2)(1+x^2)=1-x^4$$
So, the $n(\ge1)$th partial product will be $$\dfrac{1-\dfrac1{3^{2^n}}}{1-\dfrac13}$$
A: In general, you can examine that
$$(1+a)(1+a^2)(1+a^4)\cdots (1+a^{2^{k-1}}) = \sum_{i=0}^{2^k-1}a^i.$$
Its proof comes from by considering binary expansion of nonnegative integers less than $2^k$.
A: Denote the given product by $A(n)$.
Then multiply it by $B = 1 - \frac{1}{3}$
See what you get as a result, and try to draw some conclusions about $A(n)$ from there.    
A: First idea recursive sequence...
$$a_1=(1+\frac{1}{3})$$
$$a_n=a_{n-1}*(1+\frac{1}{3^{2^{n}}})    / \lim_{n \to \infty}$$
$$L=L*(1+0)$$
$$L=L$$ 
Usually this idea works, but in this case we need diffrent approach.
Look at this:
$$\lim_{n \to \infty}(1+\frac{1}{3^{2^{n}}})= 1 = \lim_{n \to \infty}{\frac{1}{3^{2^{n}}}}+\lim_{n \to \infty}{1}  = 1$$
This mean that after some step you will multiplying 1.

Let's see how: 
$\frac{4}{3}*\frac{10}{9}*\frac{82}{81}=1,499$
$\frac{4}{3}*\frac{10}{9}*\frac{82}{81}*\frac{6562}{6561}=1,499$

You only have to see that first, second and third factors are meter to result.
