#### Didn’t find what you are looking for?

Question # 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}

Series
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).
$$\sum_{n=2}^\infty\frac{5^n}{12^n}$$ 2020-12-22
Here, the given series is
$$\sum_{n=2}^\infty\frac{5^n}{12^n}$$
Let us expand the given series.
$$\sum_{n=2}^\infty\frac{5^n}{12^n}=\frac{5^2}{12^2}+\frac{5^3}{12^3}+\frac{5^4}{12^4}+...$$ (up to $$\infty)$$
$$=(\frac{5}{12})^2+(\frac{5}{12})^3+(\frac{5}{12})^4+...$$
We clearly see that the given series is an infinite geometric series with first term $$(\frac{5}{12})^2$$ and common ratio $$(\frac{5}{12})$$.
Again, We know that the sum of an infinite geometric series with first term a(say) and common ratio $$r(|r|\leq1)$$ is $$\frac{a}{1-r}$$
So, $$\sum_{n=2}^\infty\frac{5^n}{12^n}=\frac{\frac{5^2}{12^2}}{1-\frac{5}{12}}$$
$$=\frac{\frac{25}{144}}{\frac{7}{12}}$$
$$=\frac{25}{144}\times\frac{12}{7}$$
$$=\frac{25}{84}$$