# 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}

Use the formula for the sum of a geometric series to find the sum, or state that the series diverges.
$\sum _{n=2}^{\mathrm{\infty }}\frac{7\cdot \left(-3{\right)}^{n}}{{5}^{n}}$
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Given that:
The series is $\sum _{n=2}^{\mathrm{\infty }}\frac{7\cdot \left(-3{\right)}^{n}}{{5}^{n}}$
By using,
Geometric series test :
A geometric series is converges if and only if - 1 < r < 1 and its sum is,
$\sum _{n=1}^{\mathrm{\infty }}{a}_{1}{r}^{n-1}=\frac{{a}_{1}}{1-r}$
Consider the series,
$\sum _{n=2}^{\mathrm{\infty }}\frac{7\cdot \left(-3{\right)}^{n}}{{5}^{n}}=7\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-3{\right)}^{n}}{{5}^{n}}$
$=7\sum _{n=2}^{\mathrm{\infty }}\left(\frac{-3}{5}{\right)}^{n}$
The series $\sum _{n=2}^{\mathrm{\infty }}\left(\frac{-3}{5}{\right)}^{n}$ is the geometric series with first term $\left({a}_{1}\right)$ is $\frac{9}{25}$ and common ratio $\left(r\right)=-\frac{3}{5}$
Also, $|r|=|-\frac{3}{5}|\le 1$
Then,
By the geometric series,
The series $\sum _{n=2}^{\mathrm{\infty }}\left(\frac{-3}{5}{\right)}^{n}$ is converges and its sum,
$\sum _{n=2}^{\mathrm{\infty }}\left(\frac{-3}{5}{\right)}^{n}=\frac{\frac{9}{25}}{1-\left(-\frac{3}{5}\right)}=\frac{9\left(5\right)}{25\left(8\right)}=\frac{9}{40}$
Then,
$\sum _{n=2}^{\mathrm{\infty }}\frac{7\cdot \left(-3{\right)}^{n}}{{5}^{n}}=7\sum _{n=2}^{\mathrm{\infty }}\left(\frac{-3}{5}{\right)}^{n}=7\left(\frac{9}{40}\right)=\frac{63}{40}$
Therefore,
The series $\sum _{n=2}^{\mathrm{\infty }}\frac{7\cdot \left(-3{\right)}^{n}}{{5}^{n}}$ is converges and its sum is $\frac{63}{40}$

Jeffrey Jordon