determine whether the series 1/7 - 1/4 + 1/3 - 2/5+...+(-1)^(k + 1)k/(k + 6)+... converges absolutely…

Explanation:

Step1: Check for absolute - convergence

Consider the series of absolute values $\sum_{k = 1}^{\infty}\left|\frac{(- 1)^{k + 1}k}{k + 6}\right|=\sum_{k = 1}^{\infty}\frac{k}{k + 6}$. Use the Limit Comparison Test. Compare with the harmonic series $\sum_{k = 1}^{\infty}\frac{1}{1}$. Calculate the limit $\lim_{k\rightarrow\infty}\frac{\frac{k}{k + 6}}{\frac{1}{1}}=\lim_{k\rightarrow\infty}\frac{k}{k + 6}$. Divide both the numerator and denominator by $k$: $\lim_{k\rightarrow\infty}\frac{k/k}{(k + 6)/k}=\lim_{k\rightarrow\infty}\frac{1}{1+\frac{6}{k}} = 1>0$. Since the harmonic series $\sum_{k = 1}^{\infty}1$ diverges, the series of absolute values $\sum_{k = 1}^{\infty}\frac{k}{k + 6}$ diverges.

Step2: Check for conditional - convergence

Let $a_k=\frac{(-1)^{k + 1}k}{k + 6}$. First, find $\lim_{k\rightarrow\infty}a_k=\lim_{k\rightarrow\infty}\frac{(-1)^{k + 1}k}{k + 6}$. Divide numerator and denominator by $k$: $\lim_{k\rightarrow\infty}\frac{(-1)^{k + 1}}{1+\frac{6}{k}}$. The limit does not exist as $k\rightarrow\infty$ because of the $(-1)^{k + 1}$ term. By the Divergence Test (if $\lim_{k\rightarrow\infty}a_k\neq0$, then $\sum_{k = 1}^{\infty}a_k$ diverges), the series $\sum_{k = 1}^{\infty}\frac{(-1)^{k + 1}k}{k + 6}$ diverges.

Answer:

The series diverges.