use the ratio test to determine whether the series is convergent or divergent. ∑k = 1 to ∞ 5ke^(-k)…

Explanation:

Step1: Identify (a_k)

Given the series (\sum_{k = 1}^{\infty}5ke^{-k}), we have (a_k=5ke^{-k}).

Step2: Find (a_{k + 1})

Replace (k) with (k + 1) in (a_k), so (a_{k+1}=5(k + 1)e^{-(k + 1)}=\frac{5(k + 1)}{e^{k+1}}).

Step3: Calculate (\left|\frac{a_{k + 1}}{a_k}\right|)

[ \begin{align*} \left|\frac{a_{k + 1}}{a_k}\right|&=\left|\frac{\frac{5(k + 1)}{e^{k+1}}}{\frac{5k}{e^{k}}}\right|\ &=\left|\frac{5(k + 1)}{e^{k+1}}\times\frac{e^{k}}{5k}\right|\ &=\left|\frac{k + 1}{k}\times\frac{1}{e}\right|\ &=\frac{k + 1}{k}\times\frac{1}{e} \end{align*} ]

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

[ \begin{align*} \lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|&=\lim_{k\rightarrow\infty}\frac{k + 1}{k}\times\frac{1}{e}\ &=\lim_{k\rightarrow\infty}\left(1+\frac{1}{k}\right)\times\frac{1}{e}\ &=1\times\frac{1}{e}=\frac{1}{e}\approx0.368 \end{align*} ]

Answer:

(a_k = 5ke^{-k}); (\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|=\frac{1}{e}); Since (\lim_{k\rightarrow\infty}\left|\frac{a_{k + 1}}{a_k}\right|<1), the series is convergent.