# Use the formula for the sum of a geometric series to find the sum. sum_{n=4}^infty(-frac49)^n

Use the formula for the sum of a geometric series to find the sum.
$\sum _{n=4}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}$
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Pohanginah
Given series:
$\sum _{n=4}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}$
For the given series the ratio of (n+1)-th term to n-th term is constant which is $-\frac{4}{9}$
Hence, the terms of the given series are in geometric progression.
Write the given series in standard geometric series form.
That is,
$\sum _{n=4}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}=\left(-\frac{4}{9}{\right)}^{-4}+\left(-\frac{4}{9}{\right)}^{-3}+\left(-\frac{4}{9}{\right)}^{-2}+\left(-\frac{4}{9}{\right)}^{-1}+\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}$
$\sum _{n=4}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}=\frac{4365}{256}+\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}$
The general geometric series,
$\sum _{n=0}^{\mathrm{\infty }}a{r}^{n}$, where a is the first term and r is the common ratio
converges to $\frac{a}{1-r}$, for -1<r<1
Otherwise it diverges.
Compare the series $\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}$ with the general geometric series, to get
$a=1,r=-\frac{4}{9}\int \left(-1,1\right)$
Hence, $\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}=\frac{1}{1-\left(-\frac{4}{9}\right)}$
$=\frac{1}{1+\frac{4}{9}}$
$=\frac{1}{\left(\frac{13}{9}\right)}$
$=\frac{9}{13}$
Plug $\sum _{n=0}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}=\frac{9}{13}$ in equation (1), to get
$\sum _{n=-4}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}=\frac{4365}{256}+\frac{9}{13}$
$=\frac{59049}{3328}$
$\approx 17.7431$
Thus,
$\sum _{n=-4}^{\mathrm{\infty }}\left(-\frac{4}{9}{\right)}^{n}=17.7431$
Jeffrey Jordon