use the root test to determine whether the series convergent or diverge. (sum_{n = 1}^{infty}(\frac{-9n}{n +…

use the root test to determine whether the series convergent or diverge. (sum_{n = 1}^{infty}(\frac{-9n}{n + 1})^{3n}) identify (a_{n}). evaluate the following limit. (lim_{n\rightarrowinfty}sqrtn{vert a_{n}vert}) since (lim_{n\rightarrowinfty}sqrtn{vert a_{n}vert})? 1, select. need help? read it submit answer

use the root test to determine whether the series convergent or diverge. (sum_{n = 1}^{infty}(\frac{-9n}{n + 1})^{3n}) identify (a_{n}). evaluate the following limit. (lim_{n\rightarrowinfty}sqrtn{vert a_{n}vert}) since (lim_{n\rightarrowinfty}sqrtn{vert a_{n}vert})? 1, select. need help? read it submit answer

Answer

Explanation:

Step1: Identify (a_n)

Given the series (\sum_{n = 1}^{\infty}\left(\frac{-9n}{n + 1}\right)^{3n}), we have (a_n=\left(\frac{-9n}{n + 1}\right)^{3n}).

Step2: Calculate (\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_n\vert})

First, find (\vert a_n\vert=\left(\frac{9n}{n + 1}\right)^{3n}). Then (\sqrt[n]{\vert a_n\vert}=\left(\frac{9n}{n + 1}\right)^3). We simplify (\frac{9n}{n + 1}=\frac{9}{1+\frac{1}{n}}). So (\lim_{n\rightarrow\infty}\left(\frac{9n}{n + 1}\right)^3=\lim_{n\rightarrow\infty}\left(\frac{9}{1+\frac{1}{n}}\right)^3). As (n\rightarrow\infty), (\frac{1}{n}\rightarrow0). Then (\lim_{n\rightarrow\infty}\left(\frac{9}{1+\frac{1}{n}}\right)^3 = 9^3=729).

Step3: Apply Root - Test

The Root - Test states that if (\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_n\vert}=L), if (L<1), the series (\sum_{n = 1}^{\infty}a_n) is absolutely convergent (hence convergent); if (L > 1) or (L=\infty), the series (\sum_{n = 1}^{\infty}a_n) is divergent; if (L = 1), the Root - Test is inconclusive. Since (L = 729>1), the series is divergent.

Answer:

  1. (a_n=\left(\frac{-9n}{n + 1}\right)^{3n})
  2. (\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_n\vert}=729)
  3. Since (\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_n\vert}>1), the series is divergent.