Geometric series sum $\sum_{n=0}^\infty\frac{a}{(1+x)^n}$

Question

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: $$r=(1+x)^{-n}\Rightarrow a(1-(1+x))^{-n}=a/n$$

Answer 1

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 $\frac{a(x+1)}{x}$