determine whether the series (sum_{m = 2}^{infty}\frac{6}{5^{m}}) converges or diverges. if it converges…

determine whether the series (sum_{m = 2}^{infty}\frac{6}{5^{m}}) converges or diverges. if it converges, find its sum\nselect the correct choice below and, if necessary, fill in the answer box within your choice\na. the series converges because it is a geometric series with (|r|lt1). the sum of the series is\n(simplify your answer)\nb. the series diverges because (lim_{n\rightarrowinfty}\frac{6}{5^{m}}\neq0) or fails to exist\nc. the series diverges because it is a geometric series with (|r|geq1)\nd. the series converges because (lim_{n\rightarrowinfty}\frac{6}{5^{m}} = 0). the sum of the series is\n(simplify your answer)

determine whether the series (sum_{m = 2}^{infty}\frac{6}{5^{m}}) converges or diverges. if it converges, find its sum\nselect the correct choice below and, if necessary, fill in the answer box within your choice\na. the series converges because it is a geometric series with (|r|lt1). the sum of the series is\n(simplify your answer)\nb. the series diverges because (lim_{n\rightarrowinfty}\frac{6}{5^{m}}\neq0) or fails to exist\nc. the series diverges because it is a geometric series with (|r|geq1)\nd. the series converges because (lim_{n\rightarrowinfty}\frac{6}{5^{m}} = 0). the sum of the series is\n(simplify your answer)

Answer

Explanation:

Step1: Identify the form of the geometric series

A geometric series has the form (\sum_{n = k}^{\infty}ar^{n}). For the series (\sum_{m = 2}^{\infty}\frac{6}{5^{m}}), we can rewrite it as (\sum_{m = 2}^{\infty}6\times(\frac{1}{5})^{m}). Here, (a = 6\times(\frac{1}{5})^{2}=\frac{6}{25}) and (r=\frac{1}{5}). Since (|r|=\left|\frac{1}{5}\right|=\frac{1}{5}<1), the series converges.

Step2: Use the formula for the sum of an infinite geometric series

The formula for the sum of an infinite geometric series (\sum_{n = k}^{\infty}ar^{n}) is (S=\frac{a}{1 - r}) (where (|r|<1)). Substituting (a=\frac{6}{25}) and (r = \frac{1}{5}) into the formula: [ \begin{align*} S&=\frac{\frac{6}{25}}{1-\frac{1}{5}}\ &=\frac{\frac{6}{25}}{\frac{4}{5}}\ &=\frac{6}{25}\times\frac{5}{4}\ &=\frac{3}{10} \end{align*} ]

Answer:

A. The series converges because it is a geometric series with (|r|<1). The sum of the series is (\frac{3}{10})