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

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 $\frac32$.
