# Given the geometric series sum_{n=2}^inftyfrac{7(3)^n}{(-5)^n} a) Find r, the common ratio. b) Determine if the series converges or diverges. c) If it converges, find the limit.

Given the geometric series $\sum _{n=2}^{\mathrm{\infty }}\frac{7\left(3{\right)}^{n}}{\left(-5{\right)}^{n}}$
a) Find r, the common ratio.
b) Determine if the series converges or diverges.
c) If it converges, find the limit.
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Velsenw

Given geometric series is $\sum _{n=2}^{\mathrm{\infty }}\frac{7\left(3{\right)}^{n}}{\left(-5{\right)}^{n}}$
(a) To find the common ratio:
Consider the series, $\sum _{n=2}^{\mathrm{\infty }}\frac{7\left(3{\right)}^{n}}{\left(-5{\right)}^{n}}$
First term is $7×\left(-\frac{3}{5}{\right)}^{2}=\frac{63}{25}$
Second term is $7×\left(\frac{-3}{5}{\right)}^{3}=\frac{189}{125}$
Common ratio $r=\frac{\left(7×\left(-\frac{3}{5}{\right)}^{3}\right)}{\left(7×\left(-\frac{3}{5}{\right)}^{2}\right)}=-\frac{3}{5}$
(b) To determine the series is convergent or divergent:
Here common ratio is $-\frac{3}{5}$
$|r|=|-\frac{3}{5}|=\frac{3}{5}<1$, thus the series is convergent.
(c) To find the limit:
The limit is same as that of the sum of the series.
Here $\sum _{n=2}^{\mathrm{\infty }}\frac{7\left(3{\right)}^{n}}{\left(-5{\right)}^{n}}$ is a geometric series with common ratio $-\frac{3}{5}$, also which is an infinite series.
Sum of an infinite series is $\frac{a}{1-r}$, where a is the first term and r is the common ratio.
In the given series $a=7×\left(-\frac{3}{2}{\right)}^{2},r=-\frac{3}{5}$
Therefore the sum is $\frac{\left(7×\left(-\frac{3}{5}{\right)}^{2}\right)}{1-\left(-\frac{3}{5}\right)}=\frac{\left(7×\left(-\frac{3}{5}{\right)}^{2}\right)}{1+\frac{3}{5}}=\frac{63}{40}$
Thus, the limit is $\frac{63}{40}$

Jeffrey Jordon