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
asked 2021-02-15
Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
\(\frac{7}{8}-\frac{49}{64}+\frac{343}{512}-\frac{2401}{4096}+...\)

Answers (1)

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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