Answer
$\frac{1}{2}$
Work Step by Step
$S_{3}$ = $\frac{1}{2}(\frac{1}{1}-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})$ = $\frac{3}{7}$
$S_{4}$ = $\frac{1}{2}(\frac{1}{1}-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+\frac{1}{2}(\frac{1}{7}-\frac{1}{9})$ = $\frac{4}{9}$
$S_{5}$ = $\frac{1}{2}(\frac{1}{1}-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+\frac{1}{2}(\frac{1}{7}-\frac{1}{9})+\frac{1}{2}(\frac{1}{9}-\frac{1}{11})$ = $\frac{5}{11}$
$S_{N}$ = $\frac{1}{2}(\frac{1}{1}-\frac{1}{3})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{5}-\frac{1}{7})+\frac{1}{2}(\frac{1}{7}-\frac{1}{9})+\frac{1}{2}(\frac{1}{9}-\frac{1}{11})+...+\frac{1}{2}(\frac{1}{2N-1}-\frac{1}{2N+1})$ = $\frac{1}{2}(1-\frac{1}{2N+1})$
$S$ = $\lim\limits_{N \to \infty}S_{N}$ = $\lim\limits_{N \to \infty}\frac{1}{2}(1-\frac{1}{2N+1})$ = $\frac{1}{2}$
