 # Use the formula for the sum of a geometric series to find the sum or state that the series diverges (enter DIV for a divergent series). sum_{n=2}^inftyfrac{5^n}{12^n} texelaare 2020-12-21 Answered
Use the formula for the sum of a geometric series to find the sum or state that the series diverges (enter DIV for a divergent series).
$\sum _{n=2}^{\mathrm{\infty }}\frac{{5}^{n}}{{12}^{n}}$
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Here, the given series is
$\sum _{n=2}^{\mathrm{\infty }}\frac{{5}^{n}}{{12}^{n}}$
Let us expand the given series.
$\sum _{n=2}^{\mathrm{\infty }}\frac{{5}^{n}}{{12}^{n}}=\frac{{5}^{2}}{{12}^{2}}+\frac{{5}^{3}}{{12}^{3}}+\frac{{5}^{4}}{{12}^{4}}+...$ (up to $\mathrm{\infty }\right)$
$=\left(\frac{5}{12}{\right)}^{2}+\left(\frac{5}{12}{\right)}^{3}+\left(\frac{5}{12}{\right)}^{4}+...$
We clearly see that the given series is an infinite geometric series with first term $\left(\frac{5}{12}{\right)}^{2}$ and common ratio $\left(\frac{5}{12}\right)$.
Again, We know that the sum of an infinite geometric series with first term a(say) and common ratio $r\left(|r|\le 1\right)$ is $\frac{a}{1-r}$
So, $\sum _{n=2}^{\mathrm{\infty }}\frac{{5}^{n}}{{12}^{n}}=\frac{\frac{{5}^{2}}{{12}^{2}}}{1-\frac{5}{12}}$
$=\frac{\frac{25}{144}}{\frac{7}{12}}$
$=\frac{25}{144}×\frac{12}{7}$
$=\frac{25}{84}$
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