! Extra }, or forgotten $.
l.54 ... + \frac{1}{4} + \frac{1}{8} = \frac{7}[8}
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$\sum^\infty_{n=1} \frac{1}{2^n}$ \\
$s_{1} = \frac{1}{2}  s_2 = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ \\
$s_{3} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}[8}$