Question

Given that $x>0$, then find the sum of $\sum _{n=1}^{n=\infty }{\left(\frac{x}{x+1}\right)}^{n-1}$

Answer 1

$\sum _{n=1}^{\infty }{\left(\frac{x}{x+1}\right)}^{n-1}$

$=1+\frac{x}{x+1}+{\left(\frac{x}{x+1}\right)}^2+{\left(\frac{x}{x+1}\right)}^3+{\left(\frac{x}{x+1}\right)}^4+...\infty$

$=\frac{1}{1-\frac{x}{x+1}}$ since it is a G.P with $a=1$ and $r=\frac{x}{x+1}$

$=\frac{x+1}{x+1-x}$

$=x+1$