Express the following product as a single fraction: $(1+\frac{1}{3})(1+\frac{1}{9})(1+\frac{1}{81})\cdots$

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I'm having difficulty with this problem:

What i did was:

I rewrote the $1$ as $\frac{3}{3}$

here is what i rewrote the whole product as:

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

but how would i proceed after this?

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HINT: Multiply the expression by $\left(1-\frac13\right)$ and do a lot of simplifying.

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Recall the difference of squares formula $$a^2 - b^2 = (a+b)(a-b).$$ With the special case $a = 1$, $b = x^n$, we would get $$1 - x^{2n} = (1-x^n)(1+x^n).$$ What does this suggest?

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Show that for $|x|<1$ that $$\prod_{n=0}^\infty(1+x^{2^n})=\sum_{n=0}^\infty x^n.$$

Then use that $$\sum_{n=0}^\infty x^n=\frac 1{1-x}.$$

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Write it in base-3 notation. The factors are $$1.1,\; 1.01,\;1.0001,\;1.00000001\,,...,\;1.(2^n-1\text{ zeros here})1,...\;.$$They are easy to multiply: you get successively$$1.1,\; 1.111,\;1.1111111\,,...,\;1.(2^n-1\text{ ones here}),...\;.$$The limit is $1.111...$, which you probably recognize as the limit of the geometric series with first term $1$ and common ratio $\frac13$, namely $\frac32$.