determine whether the series ∑(3k + k²)/8 converges or diverges. options: diverges, converges, cannot be…

Explanation:

Step1: Analyze the first - series $\sum_{k = 1}^{\infty}\frac{3k}{8 + k^{2}}$

Use the Limit Comparison Test. Compare with the series $\sum_{k=1}^{\infty}\frac{1}{k}$. Calculate the limit $\lim_{k\rightarrow\infty}\frac{\frac{3k}{8 + k^{2}}}{\frac{1}{k}}=\lim_{k\rightarrow\infty}\frac{3k^{2}}{8 + k^{2}}$.

Step2: Evaluate the limit

Divide both the numerator and denominator by $k^{2}$: $\lim_{k\rightarrow\infty}\frac{3k^{2}/k^{2}}{(8 + k^{2})/k^{2}}=\lim_{k\rightarrow\infty}\frac{3}{\frac{8}{k^{2}}+1}=3$. Since $\sum_{k = 1}^{\infty}\frac{1}{k}$ is a harmonic series (divergent) and the limit is a positive - finite number ($L = 3>0$), the series $\sum_{k = 1}^{\infty}\frac{3k}{8 + k^{2}}$ diverges.

Step3: Analyze the second - series $\sum_{k = 1}^{\infty}\frac{1}{1 + 4k}$

Use the Limit Comparison Test. Compare with the series $\sum_{k=1}^{\infty}\frac{1}{k}$. Calculate the limit $\lim_{k\rightarrow\infty}\frac{\frac{1}{1 + 4k}}{\frac{1}{k}}=\lim_{k\rightarrow\infty}\frac{k}{1 + 4k}$.

Step4: Evaluate the limit

Divide both the numerator and denominator by $k$: $\lim_{k\rightarrow\infty}\frac{k/k}{(1 + 4k)/k}=\lim_{k\rightarrow\infty}\frac{1}{\frac{1}{k}+4}=\frac{1}{4}$. Since $\sum_{k = 1}^{\infty}\frac{1}{k}$ is a harmonic series (divergent) and the limit is a positive - finite number ($L=\frac{1}{4}>0$), the series $\sum_{k = 1}^{\infty}\frac{1}{1 + 4k}$ diverges.

Answer:

The first series $\sum_{k = 1}^{\infty}\frac{3k}{8 + k^{2}}$ diverges; the second series $\sum_{k = 1}^{\infty}\frac{1}{1 + 4k}$ diverges.