determine if the series converges or diverges. use any method, and give a reason for your answer. \n∑(n = 3…

determine if the series converges or diverges. use any method, and give a reason for your answer. \n∑(n = 3 to ∞) (n - 3)/(n9^n)\ndoes the series converge or diverge? select the correct choice below and, if necessary, fill in the answer box within your choice.\na. because (n - 3)/(n9^n) ≤ 1/9^n and ∑(n = 3 to ∞) 1/9^n converges, the series converges by the direct comparison test.\nb. since ∫(n - 3)/(n9^n) = , the given series diverges by the integral test.\nc. because (n - 3)/(n9^n) ≤ 1/n and ∑(n = 3 to ∞) 1/n converges, the series converges by the direct comparison test.\nd. since lim(n→∞) (n - 3)/(n9^n) does not exist and is not ±∞, the given series diverges by the nth - term test for divergence.\ne. since lim(n→∞) (n - 3)/(n9^n) = , the given series diverges by the nth - term test for divergence.\nf. because (n - 3)/(n9^n) ≥ 1/n and ∑(n = 3 to ∞) 1/n diverges, the series diverges by the direct comparison test.

determine if the series converges or diverges. use any method, and give a reason for your answer. \n∑(n = 3 to ∞) (n - 3)/(n9^n)\ndoes the series converge or diverge? select the correct choice below and, if necessary, fill in the answer box within your choice.\na. because (n - 3)/(n9^n) ≤ 1/9^n and ∑(n = 3 to ∞) 1/9^n converges, the series converges by the direct comparison test.\nb. since ∫(n - 3)/(n9^n) = , the given series diverges by the integral test.\nc. because (n - 3)/(n9^n) ≤ 1/n and ∑(n = 3 to ∞) 1/n converges, the series converges by the direct comparison test.\nd. since lim(n→∞) (n - 3)/(n9^n) does not exist and is not ±∞, the given series diverges by the nth - term test for divergence.\ne. since lim(n→∞) (n - 3)/(n9^n) = , the given series diverges by the nth - term test for divergence.\nf. because (n - 3)/(n9^n) ≥ 1/n and ∑(n = 3 to ∞) 1/n diverges, the series diverges by the direct comparison test.

Answer

Explanation:

Step1: Analyze the comparison for Direct - Comparison Test

For (n\geq3), we have (\frac{n - 3}{n9^{n}}=\frac{1}{9^{n}}-\frac{3}{n9^{n}}). And (\frac{n - 3}{n9^{n}}\leq\frac{1}{9^{n}}) since (\frac{3}{n9^{n}}\geq0) for (n\geq3).

Step2: Recall the geometric - series formula

The series (\sum_{n = 3}^{\infty}\frac{1}{9^{n}}) is a geometric series with common ratio (r=\frac{1}{9}). The sum of an infinite geometric series (\sum_{n = k}^{\infty}ar^{n}) (where (|r|\lt1)) is given by (\frac{ar^{k}}{1 - r}). For the series (\sum_{n=3}^{\infty}\frac{1}{9^{n}}), (a = \frac{1}{9^{3}}) and (r=\frac{1}{9}), and since (|r|=\frac{1}{9}\lt1), the series (\sum_{n = 3}^{\infty}\frac{1}{9^{n}}) converges.

Step3: Apply the Direct - Comparison Test

The Direct - Comparison Test states that if (0\leq a_{n}\leq b_{n}) for all (n\geq N) (in this case (N = 3)) and (\sum_{n = N}^{\infty}b_{n}) converges, then (\sum_{n = N}^{\infty}a_{n}) converges. Here (a_{n}=\frac{n - 3}{n9^{n}}) and (b_{n}=\frac{1}{9^{n}}).

Answer:

A. Because (\frac{n - 3}{n9^{n}}\leq\frac{1}{9^{n}}) and (\sum_{n = 3}^{\infty}\frac{1}{9^{n}}) converges, the series converges by the Direct Comparison Test.