use the ratio test to determine whether the series is convergent or divergent. \n sum_{n = 1}^{infty}(-1)^{n…

use the ratio test to determine whether the series is convergent or divergent. \n sum_{n = 1}^{infty}(-1)^{n - 1}\frac{6^{n}}{5^{n}n^{3}} \nidentify ( a_{n} ). \nevaluate the following limit. \n lim_{n\rightarrowinfty}left|\frac{a_{n + 1}}{a_{n}}\right| \nsince ( lim_{n\rightarrowinfty}left|\frac{a_{n + 1}}{a_{n}}\right|? 1 ), --select-- \nneed help? read it \nsubmit answer \n7. -/1.81 points details

use the ratio test to determine whether the series is convergent or divergent. \n sum_{n = 1}^{infty}(-1)^{n - 1}\frac{6^{n}}{5^{n}n^{3}} \nidentify ( a_{n} ). \nevaluate the following limit. \n lim_{n\rightarrowinfty}left|\frac{a_{n + 1}}{a_{n}}\right| \nsince ( lim_{n\rightarrowinfty}left|\frac{a_{n + 1}}{a_{n}}\right|? 1 ), --select-- \nneed help? read it \nsubmit answer \n7. -/1.81 points details

Answer

Explanation:

Step1: Identify (a_n)

Given the series (\sum_{n = 1}^{\infty}(-1)^{n - 1}\frac{6^{n}}{5^{n}n^{3}}), we have (a_n=(-1)^{n - 1}\frac{6^{n}}{5^{n}n^{3}}).

Step2: Find (a_{n + 1})

Replace (n) with (n+1) in (a_n), so (a_{n+1}=(-1)^{(n + 1)-1}\frac{6^{n+1}}{5^{n + 1}(n + 1)^{3}}=(-1)^{n}\frac{6^{n+1}}{5^{n+1}(n + 1)^{3}}).

Step3: Calculate (\left|\frac{a_{n+1}}{a_n}\right|)

[ \begin{align*} \left|\frac{a_{n + 1}}{a_n}\right|&=\left|\frac{(-1)^{n}\frac{6^{n+1}}{5^{n+1}(n + 1)^{3}}}{(-1)^{n - 1}\frac{6^{n}}{5^{n}n^{3}}}\right|\ &=\left|\frac{(-1)^{n}}{(-1)^{n - 1}}\right|\cdot\frac{6^{n+1}}{6^{n}}\cdot\frac{5^{n}}{5^{n+1}}\cdot\frac{n^{3}}{(n + 1)^{3}}\ &= 1\times6\times\frac{1}{5}\times\left(\frac{n}{n + 1}\right)^{3}\ &=\frac{6}{5}\left(\frac{n}{n + 1}\right)^{3} \end{align*} ]

Step4: Evaluate the limit (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|)

We know that (\lim_{n\rightarrow\infty}\frac{n}{n + 1}=\lim_{n\rightarrow\infty}\frac{1}{1+\frac{1}{n}} = 1). Then (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\rightarrow\infty}\frac{6}{5}\left(\frac{n}{n + 1}\right)^{3}=\frac{6}{5}(1)^{3}=\frac{6}{5})

Answer:

(a_n=(-1)^{n - 1}\frac{6^{n}}{5^{n}n^{3}}), (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{6}{5}), since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|> 1), the series diverges.