I have been reading Ian Stewart's Foundations of Mathematics and would greatly appreciate help with the following question.
$s_n = a + ar + ... a{r^n}$
$rs_n = ar + ar^2 + ... a{r^{n+1}}$
$rs_n - s_n = a(r^{n+1}-1) $
Deduce that
$ |s_n - \frac{a}{1-r}| = |\frac{r^{n+1}}{1-r}| . $
Factorize $s_n$ out: $$s_n(r-1)=a(r^{n+1}-1)$$ Divided by $r-1$: $$s_n=\frac{a(r^{n+1}-1)}{r-1}$$ Multiply numerator and denominator by $-1$: $$s_n=\frac{a(1-r^{n+1})}{1-r}$$
Edit: $$s_n-\frac{a}{1-r}=\frac{-ar^{n+1}}{1-r}$$
$$\left|s_n-\frac{a}{1-r}\right|=\left|\frac{-ar^{n+1}}{1-r}\right|=\left|\frac{ar^{n+1}}{1-r}\right|$$
Remark:
There was a mistake in the formula, $a$ should be present at the RHS of the formula.
Example:
let $a=2$, $r=\frac12$,$n=1$, We have $$s_1=2+1=3$$ $$\left|s_1-\frac{a}{1-r}\right|=\left|3-\frac{2}{1/2}\right|=\left|3-4\right|=1$$
$$\left|\frac{r^2}{1-r}\right|=\frac{1/4}{1/2}=\frac12$$
$$\left|\frac{ar^2}{1-r}\right|=1$$