use the ratio test to determine whether the series is convergent or di\n sum_{n = 1}^{infty}\frac{n!}{118^{n}…

use the ratio test to determine whether the series is convergent or di\n sum_{n = 1}^{infty}\frac{n!}{118^{n}} \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 master it\nsubmit answer

use the ratio test to determine whether the series is convergent or di\n sum_{n = 1}^{infty}\frac{n!}{118^{n}} \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 master it\nsubmit answer

Answer

Explanation:

Step1: Identify (a_n)

Given the series (\sum_{n = 1}^{\infty}\frac{n!}{118^n}), we have (a_n=\frac{n!}{118^n}).

Step2: Find (a_{n + 1})

Replace (n) with (n+1) in (a_n), so (a_{n+1}=\frac{(n + 1)!}{118^{n+1}}).

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

[ \begin{align*} \left|\frac{a_{n + 1}}{a_n}\right|&=\left|\frac{\frac{(n + 1)!}{118^{n+1}}}{\frac{n!}{118^n}}\right|\ &=\left|\frac{(n + 1)!}{118^{n+1}}\times\frac{118^n}{n!}\right|\ &=\left|\frac{(n + 1)\times n!}{118\times118^n}\times\frac{118^n}{n!}\right|\ &=\frac{n + 1}{118} \end{align*} ]

Step4: Evaluate the limit

[ \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\rightarrow\infty}\frac{n + 1}{118}\ &=\lim_{n\rightarrow\infty}\frac{n}{118}+\frac{1}{118}\ &=\infty \end{align*} ]

Step5: Determine convergence

Since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty> 1), by the Ratio - Test, the series (\sum_{n = 1}^{\infty}\frac{n!}{118^n}) diverges.

Answer:

(a_n=\frac{n!}{118^n}), (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\infty), since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|>1), the series diverges.