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 / 8^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 / 8^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}{8^n}), we have (a_n=\frac{n}{8^n}).

Step2: Find (a_{n + 1})

Replace (n) with (n+1) in (a_n), so (a_{n+1}=\frac{n + 1}{8^{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}{8^{n+1}}}{\frac{n}{8^n}}\right|\ &=\left|\frac{n + 1}{8^{n+1}}\times\frac{8^n}{n}\right|\ &=\left|\frac{n + 1}{n}\times\frac{8^n}{8^{n+1}}\right|\ &=\left|\frac{n + 1}{n}\times\frac{1}{8}\right|\ &=\frac{n + 1}{8n} \end{align*} ]

Step4: Evaluate the limit (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|)

[ \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|&=\lim_{n\rightarrow\infty}\frac{n + 1}{8n}\ &=\lim_{n\rightarrow\infty}\frac{n(1+\frac{1}{n})}{8n}\ &=\lim_{n\rightarrow\infty}\frac{1+\frac{1}{n}}{8}\ &=\frac{1}{8} \end{align*} ]

Step5: Determine convergence or divergence

Since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{8}<1), by the Ratio - Test, the series (\sum_{n = 1}^{\infty}\frac{n}{8^n}) is convergent.

Answer:

  1. (a_n=\frac{n}{8^n})
  2. (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=\frac{1}{8})
  3. Since (\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|<1), the series is convergent.