3.333333333 has no end.
For example,
$$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...$$
If you add 1/2 that is half to previous number and continue it,
you will see that the sum of those are going closer and closer to 1.
By common sense, they will never reach 1.
But,
$$S_n = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}$$
If I multiply this by 2, then we can see it equals to $$S_n$$.
$$2S_n=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+...+\frac{2}{2^n}
=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{2}{2^n^-^1}
=1+S_n-\frac{1}{2^n}$$
If we subtract $$S_n$$ from both sides,
$$s_n = 1 - \frac{1}{2^n}$$
As n is infinity, $$S_n = 1.$$
I think this proof apply to your question too.
As you said 3.33333+3.33333+3.33333 = 9.99999
But 3.333...+3.333...+3.333...=1
Hope it helped! 
