$3/3 = 1$. But what if I write it as $(1+1+1)/3$ or $1/3+1/3+1/3$? I can write $3/3$ as $(1+1+1)/3$ or $1/3+1/3+1/3$. 
Now, $1/3$ is a recurring/repeating/non-ending decimal so if we add these three, i.e. $0.3333... + 0.3333... + 0.3333...$ we will get infinitesimally close to $1$ but not $1$. 
Is there a way to show that these decimals do end and will eventually become $1$?
 A: $0.\bar 3$ is not in the process of becoming $1/3$ or anything else. If it is a meaningful expression then it is already equal to $1/3$ or not equal to $1/3.$
Your Q is not trivial. A logical foundation for $\mathbb R$ was only developed in the 19th century.
A: The decimals do not end, but that's not really a problem, since
$$0.999999999\dots = 1$$
There are several proofs of this, which you can look at yourself.
The simplest one is to say that if $x=0.99\dots$, then $10x = 9.99\dots$, and if you subtract the two equations you get $9x=9$ which means $x=1$.
Another way is to see that $$0.99\dots = 0.9 + 0.09 + 0.009 + \cdots =\\=9\cdot (0.1+0.1+0.001 + \cdots) = 9\cdot \sum_{i=1}^\infty 10^{-i} = 9\cdot \frac19 = 1$$
A: In the real numbers, which is the number system we ordinarily work in, and which is the unique number system for which decimal notation (i.e., writing numbers as arbitrary infinite sequences of decimals) works, there is no such thing as "infinitesimally close but not equal". In fact, $0.333...$ is exactly equal to $\frac13$, and
$$
1 = \frac33 = \frac13 + \frac13 + \frac13 = 0.333... + 0.333... + 0.333... = 0.999... = 1.
$$
A: $0,333333.....= 3\sum _{n=1}^{\infty}\frac{1}{10^n}=\frac{1}{3}$
