Rational as series?

Question

I was checking out a few things in the geometric series and apparently all rational numbers can be shown as a geometric series $$ \frac{p}{q} = \frac{p}{{q + 1}} + \frac{p}{{(q + 1)^2 }} + \frac{p}{{(q + 1)^3 }} + \ldots \,. $$ What is a good proof for the equation?

Answer 1

You have $$ \sum_{n=1}^\infty \frac{1}{(q+1)^n} = \frac{1}{q+1} \cdot \frac{1}{1-\frac{1}{q+1}}= \frac{1}{q} $$ using the formula for geometric series applied to $r = \frac{1}{q+1}$, recalling that $0< \frac{1}{q+1} < 1$. Therefore, $$ \sum_{n=1}^\infty \frac{p}{(q+1)^n}= \frac{p}{q} $$ as claimed.

I don't know who showed this first, but this is pretty straightforward.

Answer 2

The formula you wrote is not entirely correct. What you have is $$\tag1 \frac pq=\sum_{k=1}^\infty \left(\frac p{p+q}\right)^k. $$ This works nicely when $p,q>0$, since in this case $\left|\tfrac p{p+q}\right|<1$. The formula does not work for many negative numbers.

As for who may have come up with such formula, the geometric series has been known for millennia: the formula, at least for a finite sum, already appears in Euclid's work. This is basic enough that I doubt it ever appeared as a theorem in anybody's work.