determine if the series converges or diverges. use any method, and give a reason for your answer\n\nselect…

determine if the series converges or diverges. use any method, and give a reason for your answer\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice\n\na the series diverges because the limit used in the nth - term test is\n(type an exact answer)\n\nb because \\( \\frac { 1 } { 4 ^ { n - 1 } + 9 } \\geq \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) and \\( \\sum _ { n = 1 } ^ { \\infty } \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) diverges, the series diverges by the direct comparison test\n\nc because \\( \\frac { 1 } { 4 ^ { n - 1 } + 9 } \\leq \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) and \\( \\sum _ { n = 1 } ^ { \\infty } \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) converges the series converges by the direct comparison test

determine if the series converges or diverges. use any method, and give a reason for your answer\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice\n\na the series diverges because the limit used in the nth - term test is\n(type an exact answer)\n\nb because \\( \\frac { 1 } { 4 ^ { n - 1 } + 9 } \\geq \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) and \\( \\sum _ { n = 1 } ^ { \\infty } \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) diverges, the series diverges by the direct comparison test\n\nc because \\( \\frac { 1 } { 4 ^ { n - 1 } + 9 } \\leq \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) and \\( \\sum _ { n = 1 } ^ { \\infty } \\left( \\frac { 1 } { 4 } \\right) ^ { n - 1 } \\) converges the series converges by the direct comparison test

Answer

Explanation:

Step1: Analyze the general term of the series

We have (a_{n}=\frac{1}{4^{n - 1}+9}). Since (4^{n-1}+9>4^{n - 1}) for (n\geq1), then (\frac{1}{4^{n - 1}+9}<\frac{1}{4^{n - 1}}=\left(\frac{1}{4}\right)^{n - 1}).

Step2: Recall the formula for the sum of a geometric series

The sum of a geometric series (\sum_{n = 1}^{\infty}r^{n-1}) is given by (S=\frac{1}{1 - r}) when (|r|<1). For the series (\sum_{n=1}^{\infty}\left(\frac{1}{4}\right)^{n - 1}), where (r=\frac{1}{4}) (and (|r|=\frac{1}{4}<1)), the sum is (S=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}).

Answer:

C. Because (\frac{1}{4^{n - 1}+9}<\left(\frac{1}{4}\right)^{n - 1}) and (\sum_{n = 1}^{\infty}\left(\frac{1}{4}\right)^{n - 1}) converges, the series converges by the Direct Comparison Test.