Determine convergence or divergence of the series.
-$$\sum _ { n = 1 } ^ { \infty } \frac { 4 n + 6 } { \sqrt { 6 n ^ { 3 } + 5 n + 7 } }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

We can use the limit comparison test with the series $\sum \frac{1}{\sqrt{n}}$.
\begin{align*}
\lim_{n	o\infty}\frac{\frac{4n+6}{\sqrt{6n^3+5n+7}}}{\frac{1}{\sqrt{n}}} &= \lim_{n	o\infty} \frac{4n+6}{\sqrt{n}\sqrt{6n^3+5n+7}} \
&= \lim_{n	o\infty}\frac{4+\frac{6}{n}}{\sqrt{6+\frac{5}{n^2}+\frac{7}{n^3}}}\
&=\frac{4}{\sqrt{6}}>0.
\end{align*}
Since $\sum \frac{1}{\sqrt{n}}$ is a $p$-series with $p=\frac{1}{2}<1$ and diverges, by the limit comparison test, the given series $\sum \frac{4n+6}{\sqrt{6n^3+5n+7}}$ also diverges. Therefore, the answer is $\boxed{	extbf{(B)}}$.

Determine Convergence or Divergence of the Series ∑n=1∞4n+66n3+5n+7\sum _ { n = 1 } ^ { \infty } \frac { 4 n + 6 } { \sqrt { 6 n ^ { 3 } + 5 n + 7 } }n=1∑∞​6n3+5n+7​4n+6​

Determine Convergence or Divergence of the Series $\sum _ { n = 1 } ^ { \infty } \frac { 4 n + 6 } { \sqrt { 6 n ^ { 3 } + 5 n + 7 } }$