Solusi

$S=\large{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...}$

Karena untuk $n \in \mathbb{Z^+}$ berlaku $\large{\frac{1}{2^n}-\frac{1}{2^{n+1}}}=\frac{1}{2^{n+1}}$ maka

$S=\large{(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{2^2})+(\frac{1}{2^2}-\frac{1}{2^3})+(\frac{1}{2^3}-\frac{1}{2^4})+...}$

$S=\large{1-(\frac{1}{2}-\frac{1}{2})-(\frac{1}{2^2}-\frac{1}{2^2})-(\frac{1}{2^3}-\frac{1}{2^3})-(\frac{1}{2^4}-...}$

$S=\large{1-\cancel{(\frac{1}{2}-\frac{1}{2})}-\cancel{(\frac{1}{2^2}-\frac{1}{2^2})}-\cancel{(\frac{1}{2^3}-\frac{1}{2^3})}-\cancel{(\frac{1}{2^4}-}...}$

$S=1$