# Use the formula for the sum of a geometric series to find the sum, or state that the series diverges. frac{7}{8}-frac{49}{64}+frac{343}{512}-frac{2401}{4096}+...

Question
Series
Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
$$\frac{7}{8}-\frac{49}{64}+\frac{343}{512}-\frac{2401}{4096}+...$$

2021-02-16
Given geometrical series is $$\frac{7}{8}-\frac{49}{64}+\frac{343}{512}-\frac{2401}{4096}+...$$
First term of the series is $$a_1=\frac{7}{8}$$
Common ratio will be:
$$r=\frac{a_2}{a_1}$$
$$=\frac{(-\frac{49}{64})}{(\frac{7}{8})}$$
$$=-\frac{49}{64}\cdot\frac{8}{7}$$
$$=-\frac{7}{8}$$
Sum of infinite terms of geometric series is given as:
$$S=\frac{a_1}{1-r}$$
Therefore, sum of given geometric series will be:
$$S=\frac{\frac{7}{8}}{(1-(-\frac{7}{8}))}$$
$$=\frac{\frac{7}{8}}{(1+\frac{7}{8})}$$
$$=\frac{\frac{7}{8}}{(\frac{15}{8})}$$
$$=\frac{7}{15}$$
Hence, required sum of the given series is $$\frac{7}{15}$$

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