Geometric series sum $\sum_{n=0}^\infty\frac{a}{(1+x)^n}$ Why does this hold
$$\sum_{n=0}^\infty\frac{a}{(1+x)^n}=\frac{a(x+1)}{x}$$
?
To me it looks like $=\frac{a}{x}$ from the formula.
That is:
$$r=(1+x)^{-n}\Rightarrow a(1-(1+x))^{-n}=a/n$$

Okay, do if you put $n=0,1,2..$ in the forumula you'll get,
this series,
$a, \frac{a}{1+x}, \frac{a}{(1+x)^2}...$
with common ration $\frac{1}{1+x}$ now we can use the forumula for sum of $\infty$ GP Which is : $\frac{a}{1-r}$ here $r$ is common ratio;
so we get-$\frac{a}{1-\frac{1}{1-x}}$ which becomes $\frac{a(x+1)}{x}$
