# 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 Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
$$\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}$$ 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 $$\frac{63}{40}$$

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