Is 1 divided by 3 equal to 0.333...?

Here is a simple reasoning that $1/3=0.3333...$.

Lets denote $0.333333......$ by $x$. Then

$$x=0.33333.....$$ $$10x=3.33333...$$

Subtracting we get $9x=3$. Thus $x=\frac{3}{9}=\frac{1}{3}$.

Since $x$ was chosen as $0.3333....$ it means that $0.3333...=\frac{1}{3}$.

You can find the sum of $\frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots$ using the formula of sum of infinite geometric progression.

$$a_1 = \frac{3}{10}$$

$$r=\frac{1}{10}$$

$$\sum =\frac{a_1}{1-r}=\frac{3}{10}\times\frac{10}{9} =\frac{1}{3}$$

The problematic part of the question is "no matter how many ones you add, 0.111... will never equal precisely 1/9."

In this (imprecise) context $0.111\ldots$ is an infinite sequence of ones; the sequence of ones does not terminate, so there is no place at which to add another one; each one is already followed by another one. Thus, $10\times0.111\ldots=1.111\ldots$ is precise. Therefore, $9\times0.111\ldots=1.000\ldots=1$ is precise, and $0.111\ldots=1/9$.

I say "imprecise" because we also say $\pi=3.14159\dots$ where ... there means an unspecified sequence of digits following. A more precise way of writing what, in the context of this question, we mean by $0.111\dots$ is $0.\overline{1}$ where the group of digits under the bar is to be repeated without end.

In this question, $0.333\ldots=0.\overline{3}$, and just as above, $10\times0.\overline{3}=3.\overline{3}$, and therefore, $9\times0.\overline{3}=3.\overline{0}=3$, which means $0.\overline{3}=3/9=1/3$.