How to write $a/(c+1)$ as $(a/c) + \cdots$?

Question

I am looking for a way to rewrite

$$ \frac{a}{c+1} $$ as

$$ \frac{a}{c} + \cdots $$

in other words, move the $+ 1$ part outside of the denominator.

(a and c are real numbers)

Is this possible? If so, how?

Answer 1

We go a bit farther and show that $\tfrac{a}{c+1}$ and $\tfrac{a}{c}$ are algebraically independent whenever $a$ and $c$ are.

Assume $a$ and $c$ are algebraically independent: that is, for all polynomials $f$ with integer coefficients, $$f(a,c)=0\Rightarrow f = 0\text{.}$$ Suppose we had a polynomial relation $$p(\tfrac{a}{c+1},\tfrac{a}{c})=0$$ between $\tfrac{a}{c+1}$ and $\tfrac{a}{c}$. Then since $$p(\tfrac{a}{c+1},\tfrac{a}{c})=(c+1)^{-i}c^{-j}q(a,c)$$ for some natural numbers $i$, $j$ and polynomial $q$, we would have a relation $$q(a,c)=0\text{,}$$ implying that $q=0$. Therefore $p=0$ as well, i.e., $$p(\tfrac{a}{c+1},\tfrac{a}{c})=0\Rightarrow p =0\text{.}$$ QED

Answer 2

For something slightly different, we have \begin{align} \frac{a}{1 + c} &= \frac{a}{c} \left( \frac{1}{1 + \frac{1}{c}}\right) = \frac{a}{c} \left( \sum_{n = 0}^\infty \frac{(-1)^n}{c^n} \right) \\&= \frac{a}{c}\left(1 - \frac{1}{c} + \frac{1}{c^2} + \cdots \right) = \frac{a}{c} - \frac{a}{c^2} + \frac{a}{c^3} + \cdots \end{align} This series will converge as long as $|c| > 1$.