use the ratio test to determine whether the series is convergent or divergent.\n∑(n = 1 to ∞) n /…

use the ratio test to determine whether the series is convergent or divergent.\n∑(n = 1 to ∞) n / 6^n\nidentify a_n.\nevaluate the following limit.\nlim(n → ∞) |a_n + 1 / a_n|\nsince lim(n → ∞) |a_n + 1 / a_n|? 1, select-\nneed help? read it watch it\nsubmit answer

use the ratio test to determine whether the series is convergent or divergent.\n∑(n = 1 to ∞) n / 6^n\nidentify a_n.\nevaluate the following limit.\nlim(n → ∞) |a_n + 1 / a_n|\nsince lim(n → ∞) |a_n + 1 / a_n|? 1, select-\nneed help? read it watch it\nsubmit answer

Answer

Explanation:

Step1: Identify (a_n)

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

Step2: Find (a_{n + 1})

Replace (n) with (n+1) in (a_n), so (a_{n+1}=\frac{n + 1}{6^{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}{6^{n+1}}}{\frac{n}{6^n}}\right|\ &=\left|\frac{n + 1}{6^{n+1}}\times\frac{6^n}{n}\right|\ &=\left|\frac{n + 1}{6n}\right|\ &=\frac{n+1}{6n}\ &=\frac{1}{6}\left(1+\frac{1}{n}\right) \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{1}{6}\left(1+\frac{1}{n}\right)\ &=\frac{1}{6}\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)\ &=\frac{1}{6}(1 + 0)\ &=\frac{1}{6} \end{align*} ]

Answer:

(a_n=\frac{n}{6^n}), (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{6}), since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{6}<1), the series is convergent.