use the ratio test to determine whether the series is convergent or divergent. ∑k = 1 to ∞ 3/k! identify…

use the ratio test to determine whether the series is convergent or divergent. ∑k = 1 to ∞ 3/k! identify a_k. 3/k! evaluate the following limit. lim k → ∞ |a_k + 1/a_k| since lim k → ∞ |a_k + 1/a_k|? 1, select need help? read it submit answer

use the ratio test to determine whether the series is convergent or divergent. ∑k = 1 to ∞ 3/k! identify a_k. 3/k! evaluate the following limit. lim k → ∞ |a_k + 1/a_k| since lim k → ∞ |a_k + 1/a_k|? 1, select need help? read it submit answer

Answer

Explanation:

Step1: Find (a_{k + 1})

Given (a_{k}=\frac{3}{k!}), then (a_{k + 1}=\frac{3}{(k + 1)!})

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

[ \begin{align*} \left|\frac{a_{k + 1}}{a_{k}}\right|&=\left|\frac{\frac{3}{(k + 1)!}}{\frac{3}{k!}}\right|\ &=\left|\frac{3}{(k + 1)!}\times\frac{k!}{3}\right|\ &=\left|\frac{k!}{(k + 1)!}\right|\ &=\left|\frac{k!}{(k+1)\times k!}\right|\ &=\frac{1}{k + 1} \end{align*} ]

Step3: Evaluate the limit

[ \begin{align*} \lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|&=\lim_{k\rightarrow\infty}\frac{1}{k + 1}\ &= 0 \end{align*} ]

Answer:

  1. Since (\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|=0<1), the series (\sum_{k = 1}^{\infty}\frac{3}{k!}) is convergent.