Question

Determine if the series converges or diverges. Use any method, and give a reason for your answer. $$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3 - 6k + 24}}$$ Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The series converges because it is a geometric series with r = B. The series diverges because it is a p-series with p = C. Because $$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3 - 6k + 24}} \le \sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3}}$$ and $$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3}}$$ converges, the series converges by the Direct Comparison Test. D. The series diverges because the limit found in the nth-Term Test is E. The series diverges per the Integral Test because $$\int_{4}^{\infty} \frac{1}{\sqrt{k^3 - 6k + 24}} dx = $$ F. Since $$\lim_{k \to \infty} \frac{a_k}{b_k} = 1$$, 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 $$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3}}$$ converge Limit Comparison Test.

Name: determine if the series converges or diverges use any method and give a reason for your answer sumk4infty frac1sqrtk3 6k 24 select the correct choice below and if necessary fill in the answ 03505
Uploaded: 2020-09-23T14:28:54-08:00
Duration: 58 s
Channel: Linh Vu
Description: determine if the series converges or diverges use any method and give a reason for your answer sumk4infty frac1sqrtk3 6k 24 select the correct choice below and if necessary fill in the answ 03505

          Determine if the series converges or diverges. Use any method, and give a reason for your answer.
$$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3 - 6k + 24}}$$
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The series converges because it is a geometric series with r = 
B. The series diverges because it is a p-series with p = 
C. Because $$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3 - 6k + 24}} \le \sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3}}$$ and $$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3}}$$ converges, the series converges by the Direct Comparison Test.
D. The series diverges because the limit found in the nth-Term Test is 
E. The series diverges per the Integral Test because $$\int_{4}^{\infty} \frac{1}{\sqrt{k^3 - 6k + 24}} dx = $$
F. Since $$\lim_{k \to \infty} \frac{a_k}{b_k} = 1$$, 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 $$\sum_{k=4}^{\infty} \frac{1}{\sqrt{k^3}}$$ converge
Limit Comparison Test.

$determine if the series converges or diverges use any method and give a reason for your answer sumk4infty frac1sqrtk3 6k 24 select the correct choice below and if necessary fill in the answ 03505$

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