# Find the sum of the series. sum_(n=1)^(infty) 7/(4^n)

Find the sum of the series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{7}{{4}^{n}}$
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kamphundg4
$\sum _{n=1}^{\mathrm{\infty }}=7\left(\frac{1}{4}+\left(\frac{1}{4}{\right)}^{2}+\left(\frac{1}{4}{\right)}^{3}+...+\left(\frac{1}{4}{\right)}^{N}+...\right)$
$=\frac{7}{4}\left(1+\left(\frac{1}{4}{\right)}^{2}+\left(\frac{1}{4}{\right)}^{3}+....\right)$ this is a geometric series
$r=\frac{1}{4}<1$
and

$\underset{n\to \mathrm{\infty }}{lim}\frac{7}{4}\left(\frac{1-\left(\frac{1}{4}{\right)}^{n}}{1-\frac{1}{4}}\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{7}{4}\ast \frac{1}{\frac{3}{4}}=\frac{7.4}{4.3}=\frac{7}{3}$