Question

Can anyone help???

Answer 1

$\sum\limits^{\infty}_{i = 1}{8(\frac{5}{6})^{i - 1}} = 8(\frac{5}{6})^{1 - 1} + 8(\frac{5}{6})^{2 - 1} + 8(\frac{5}{6})^{3 - 1} +...$
$\sum\limits^{\infty}_{i = 1}{8(\frac{5}{6})^{i - 1}} = 8(\frac{5}{6})^{0} + 8(\frac{5}{6})^{1} + 8\frac{5}{6})^{2} +...$
$\sum\limits^{\infty}_{i = 1} = 8(1) + 8(\frac{5}{6}) + 8(\frac{25}{36})+...$
$\sum\limits^{\infty}_{i = 1} = 8 + 6\frac{2}{3} + 5\frac{5}{9} +...$
$\sum\limits^{\infty}_{i = 1} = 20\frac{2}{9} +...$

The series is a divergent geometric series.