Use the formula for the sum of a geometric series to find the sum, or state that the series diverges. sum_{n=2}^inftyfrac{7cdot(-3)^n}{5^n}

Question
Series
asked 2021-01-31
Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
\(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}\)

Answers (1)

2021-02-01
Given that:
The series is \(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}\)
By using,
Geometric series test :
A geometric series \(\sum_{n=0}^\infty a_1r^n\text{ or }\sum_{n=1}^\infty a_1r^{n-1}\) is converges if and only if - 1 < r < 1 and its sum is,
\(\sum_{n=1}^\infty a_1r^{n-1}=\frac{a_1}{1-r}\)
Consider the series,
\(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}=7\sum_{n=2}^\infty\frac{(-3)^n}{5^n}\)
\(=7\sum_{n=2}^\infty(\frac{-3}{5})^n\)
The series \(\sum_{n=2}^\infty(\frac{-3}{5})^n\) is the geometric series with first term \((a_1)\) is \(\frac{9}{25}\) and common ratio \((r)=-\frac35\)
Also, \(|r|=|-\frac35|\leq1\)
Then,
By the geometric series,
The series \(\sum_{n=2}^\infty(\frac{-3}{5})^n\) is converges and its sum,
\(\sum_{n=2}^\infty(\frac{-3}{5})^n=\frac{\frac{9}{25}}{1-(-\frac35)}=\frac{9(5)}{25(8)}=\frac{9}{40}\)
Then,
\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}=7\sum_{n=2}^\infty(\frac{-3}{5})^n=7(\frac{9}{40})=\frac{63}{40}\)
Therefore,
The series \(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}\) is converges and its sum is \(\frac{63}{40}\)
0

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