determine whether the series ∑((-1)^k*e^(-6k))/5 converges absolutely, converges conditionally or diverges…

Explanation:

Step1: Consider absolute - value series

We first consider the series of absolute values of (\sum_{k = 1}^{\infty}\frac{(- 1)^{k}ke^{-6k}}{5}), which is (\sum_{k = 1}^{\infty}\frac{ke^{-6k}}{5}=\frac{1}{5}\sum_{k = 1}^{\infty}ke^{-6k}).

Step2: Apply the ratio - test

Let (a_{k}=ke^{-6k}). Then (a_{k + 1}=(k + 1)e^{-6(k+1)}). Calculate the ratio (\left|\frac{a_{k + 1}}{a_{k}}\right|=\left|\frac{(k + 1)e^{-6(k + 1)}}{ke^{-6k}}\right|=\left|\frac{k + 1}{k}\cdot e^{-6}\right|). As (k\to\infty), (\lim_{k\to\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|=\lim_{k\to\infty}\left(1+\frac{1}{k}\right)e^{-6}=e^{-6}\approx0.0025\lt1). Since (\lim_{k\to\infty}\left|\frac{a_{k + 1}}{a_{k}}\right|\lt1) for the series (\sum_{k = 1}^{\infty}a_{k}), by the ratio - test, the series (\sum_{k = 1}^{\infty}\frac{ke^{-6k}}{5}) converges.

Step3: Conclusion on convergence type

Since the series of absolute values (\sum_{k = 1}^{\infty}\left|\frac{(-1)^{k}ke^{-6k}}{5}\right|) converges, the original series (\sum_{k = 1}^{\infty}\frac{(-1)^{k}ke^{-6k}}{5}) converges absolutely.

Answer:

converges absolutely