dmm, 2 months ago @johncarlosbaez It's true, but the problem (back when I first saw it) was posed in terms of (\frac{a}{b}) so I worked that way. But you are right, I should do [S=\sum\limits_{n=1}^{\infty} x^n] (which converges to (\frac{x}{1-x}) for (|x| < 1), similar logic...)
@johncarlosbaez It's true, but the problem (back when I first saw it) was posed in terms of (\frac{a}{b}) so I worked that way.
But you are right, I should do [S=\sum\limits_{n=1}^{\infty} x^n] (which converges to (\frac{x}{1-x}) for (|x| < 1), similar logic...)