New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
$${1\over(1-x)^2}={1\over 1-x}\cdot{1\over 1-x}=\sum_{j\geq0} x^j\cdot\sum_{k\geq0}x^k =\sum_{r\geq0} x^r\left(\sum_{j+k=r}1\right)=\sum_{r\geq0}(r+1)x^r\ .$$
For the special case $x=\dfrac12$:
If you accept that $1+x+x^2+\dotsb=\dfrac1{1-x}$, the same picture works — just move the horizontal and vertical lines. Instead of them being at $1,1\frac12,1\frac34,\dotsb,2$, you should put them at $1,1+x,1+x+x^2,\dotsb,\dfrac1{1-x}$. The area of the square is then $\left(\dfrac1{1-x}\right){}^2$.
