use the ratio test to determine if the following series converges absolutely or diverges\n sum_{n =…

use the ratio test to determine if the following series converges absolutely or diverges\n sum_{n = 1}^{infty}\frac{n^{19}}{(-10)^{n}}\nsince the limit resulting from the ratio test is (\frac{1}{10}), which is (<1) (simplify your answer )\nthe series converges absolutely\nthe ratio test is inconclusive\nthe series diverges

use the ratio test to determine if the following series converges absolutely or diverges\n sum_{n = 1}^{infty}\frac{n^{19}}{(-10)^{n}}\nsince the limit resulting from the ratio test is (\frac{1}{10}), which is (<1) (simplify your answer )\nthe series converges absolutely\nthe ratio test is inconclusive\nthe series diverges

Answer

Explanation:

Step1: Recall the Ratio Test formula

For a series (\sum_{n = 1}^{\infty}a_{n}), calculate (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|). Here (a_{n}=\frac{n^{19}}{(- 10)^{n}}), then (a_{n+1}=\frac{(n + 1)^{19}}{(-10)^{n+1}}).

Step2: Calculate (\left|\frac{a_{n + 1}}{a_{n}}\right|)

[ \begin{align*} \left|\frac{a_{n + 1}}{a_{n}}\right|&=\left|\frac{\frac{(n + 1)^{19}}{(-10)^{n+1}}}{\frac{n^{19}}{(-10)^{n}}}\right|\ &=\left|\frac{(n + 1)^{19}}{(-10)^{n+1}}\times\frac{(-10)^{n}}{n^{19}}\right|\ &=\left|\frac{(n + 1)^{19}}{n^{19}}\times\frac{-10^{n}}{10^{n+1}}\right|\ &=\frac{1}{10}\left(\frac{n + 1}{n}\right)^{19}\ &=\frac{1}{10}\left(1+\frac{1}{n}\right)^{19} \end{align*} ]

Step3: Find the limit as (n\rightarrow\infty)

(\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\frac{1}{10}\left(1+\frac{1}{n}\right)^{19}). Since (\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{19}=1) (using the fact that (\lim_{x\rightarrow\infty}(1+\frac{1}{x})^{k}=1) for any constant (k)), then (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\frac{1}{10})

Answer:

the series converges absolutely