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

Explanation:

Step1: Write the formula for the Ratio Test

For a series (\sum_{n = 1}^{\infty}a_{n}), the ratio test requires us to find (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|). Here, (a_{n}=\frac{n^{19}}{(- 10)^{n}}), and (a_{n+1}=\frac{(n + 1)^{19}}{(-10)^{n+1}}). Then (\left|\frac{a_{n + 1}}{a_{n}}\right|=\left|\frac{\frac{(n + 1)^{19}}{(-10)^{n+1}}}{\frac{n^{19}}{(-10)^{n}}}\right|).

Step2: Simplify the expression

[ \begin{align*} \left|\frac{a_{n + 1}}{a_{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

(\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)^{k}=1) for any constant (k) (by the well - known limit (\lim_{x\rightarrow\infty}(1+\frac{1}{x})^{x}=e) and for non - variable exponents (\lim_{x\rightarrow\infty}(1 + \frac{1}{x})^{k}=\left(\lim_{x\rightarrow\infty}(1+\frac{1}{x})\right)^{k})), when (k = 19), we have (\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{19}=1). So (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\frac{1}{10}\times1=\frac{1}{10}).

Answer:

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