determine if the series converges or diverges. use any method, and give a reason for your answer. \n∑k = 4…

determine if the series converges or diverges. use any method, and give a reason for your answer. \n∑k = 4 to ∞ 1/√(k³ - 6k + 24)\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the series diverges because the limit found in the nth - term test is \n b. the series diverges per the integral test because ∫4 to ∞ 1/√(k³ - 6k + 24)dx=\n c. the series converges because it is a geometric series with r =\n d. the series diverges because it is a p - series with p =\n e. because 1/√(k³ - 6k + 24)≤1/√(k³) and ∑k = 4 to ∞ 1/√(k³) converges, the series converges by the direct comparison test.\n f. since limk→∞ aₖ/bₖ = 1, where aₖ = 1/√(k³ - 6k + 24) and bₖ = 1/√(k³), both series have positive terms, and the series ∑k = 4 to ∞ 1/√(k³) converges, the given series converges by the limit comparison test.

determine if the series converges or diverges. use any method, and give a reason for your answer. \n∑k = 4 to ∞ 1/√(k³ - 6k + 24)\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the series diverges because the limit found in the nth - term test is \n b. the series diverges per the integral test because ∫4 to ∞ 1/√(k³ - 6k + 24)dx=\n c. the series converges because it is a geometric series with r =\n d. the series diverges because it is a p - series with p =\n e. because 1/√(k³ - 6k + 24)≤1/√(k³) and ∑k = 4 to ∞ 1/√(k³) converges, the series converges by the direct comparison test.\n f. since limk→∞ aₖ/bₖ = 1, where aₖ = 1/√(k³ - 6k + 24) and bₖ = 1/√(k³), both series have positive terms, and the series ∑k = 4 to ∞ 1/√(k³) converges, the given series converges by the limit comparison test.

Answer

Explanation:

Step1: Choose a comparison series

For large - $k$, the dominant term in the denominator of $a_{k}=\frac{1}{\sqrt{k^{3}-6k + 24}}$ is $\sqrt{k^{3}}$. So, we choose the comparison series $\sum_{k = 4}^{\infty}b_{k}=\sum_{k=4}^{\infty}\frac{1}{\sqrt{k^{3}}}=\sum_{k = 4}^{\infty}k^{-\frac{3}{2}}$, which is a $p$-series with $p=\frac{3}{2}>1$. A $p$-series $\sum_{k = 1}^{\infty}\frac{1}{k^{p}}$ converges if $p>1$.

Step2: Apply the Limit - Comparison Test

Calculate the limit $\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}$, where $a_{k}=\frac{1}{\sqrt{k^{3}-6k + 24}}$ and $b_{k}=\frac{1}{\sqrt{k^{3}}}$. [ \begin{align*} \lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}&=\lim_{k\rightarrow\infty}\frac{\frac{1}{\sqrt{k^{3}-6k + 24}}}{\frac{1}{\sqrt{k^{3}}}}\ &=\lim_{k\rightarrow\infty}\sqrt{\frac{k^{3}}{k^{3}-6k + 24}}\ &=\lim_{k\rightarrow\infty}\sqrt{\frac{1}{1-\frac{6}{k^{2}}+\frac{24}{k^{3}}}}\ & = 1 \end{align*} ] Since $\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}} = 1$, both series have positive terms, and $\sum_{k = 4}^{\infty}b_{k}=\sum_{k=4}^{\infty}\frac{1}{\sqrt{k^{3}}}$ converges, by the Limit - Comparison Test, the series $\sum_{k = 4}^{\infty}\frac{1}{\sqrt{k^{3}-6k + 24}}$ converges.

Answer:

F. The series converges because since $\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}} = 1$, where $a_{k}=\frac{1}{\sqrt{k^{3}-6k + 24}}$ and $b_{k}=\frac{1}{\sqrt{k^{3}}}$, both series have positive terms, and the series $\sum_{k = 4}^{\infty}\frac{1}{\sqrt{k^{3}}}$ converges, the given series converges by the Limit Comparison Test.