use the ratio test to determine if the following series converges absolutely or diverges\n sum _ { n = 1 } ^…

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

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

[ \begin{align*} \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)^{10}\ \end{align*} ] Since (\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{10}=1) (using the fact that (\lim_{x\rightarrow\infty}(1+\frac{1}{x})^{k}=1) for any real - number (k)), then (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\frac{1}{10})

Answer:

Since the limit resulting from the Ratio Test is (\frac{1}{10}), which is (<1), the series converges absolutely.