How to explain the formula for the sum of a geometric series without calculus? How to explain to a middle-school student  the notion of a geometric series without any calculus (i.e. limits)? For example I want to convince my student that
$$1 + \frac{1}{4} + \frac{1}{4^2} + \ldots + \frac{1}{4^n} = \frac{1 - (\frac{1}{4})^{n+1} }{ 1 - \frac{1}{4}}$$
at $n \to \infty$ gives 4/3?

$$
\frac{1}{3}-\frac{1}{4}=\frac{1}{12}=\frac{1}{4}\cdot\frac{1}{3}
$$
so
$$
\frac{1}{3}=\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{3}
$$
Now ask your 14 year old to plug in this expression for $1\over 3$ into itself, quite funny, bewildering and strange at first sight:
$$
\frac{1}{3}=\frac{1}{4}+\frac{1}{4}\cdot \left(\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{3}\right)=
\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^2}\frac{1}{3}
$$
Repeat two or three times, then discuss the difference
$$
\frac{1}{3}-\left(\frac{1}{4}+\frac{1}{4^2}+\dots+\frac{1}{4^n}\right)
$$
edit 
For example I want to convince my student that
this is impossible without talking about the notion a limit. How do you convince a student that $1,{1\over 2},{1 \over 3},\dots$ goes to zero? What does goes to even mean? In this situation
$$
1+\frac{1}{4}+\frac{1}{4^2}+\dots=\frac{4}{3}
$$
You cannot convince somebody that this is true without defining the meaning of these little dots on the left side.
