Use the Limit Comparison Test (if possible) to determine whether the series $$\sum _ { n = 1 } ^ { \infty } \frac { 2 } { \sqrt [ 9 ] { n ^ { 2 } + 9 } }$$
A) diverges
B) converges
C) Limit Comparison Test does not apply

Question

Quizplus · Accepted Answer

We can use the Limit Comparison Test by comparing the given series to a known convergent or divergent series. Let's choose the series $\sum \frac{1}{n^{3/2}}$, which is a known convergent p-series. 
Then we have 
\begin{align*}
\lim_{n 	o \infty} \frac{\frac{2}{\sqrt[9]{n^2+9}}}{\frac{1}{n^{3/2}}} &= \lim_{n 	o \infty} 2n^\frac{3}{2}\left(n^2 + 9ight)^{-1/9}\
&= 2\lim_{n 	o \infty} \left(\frac{n^\frac{2}{9}}{(n^2+9)^{1/9}}ight)^3 \
&= 2\lim_{n 	o \infty} \left(\frac{1}{n^{16/9}}ight) = 0
\end{align*}
Since the limit is finite and positive, and $\sum \frac{1}{n^{3/2}}$ converges, by the Limit Comparison Test, the original series $\sum \frac{2}{\sqrt[9]{n^2+9}}$ also converges.

However, we can also note that each term of the series is greater than or equal to $\frac{2}{n^{2/9}}$, and the series $\sum\frac{1}{n^{2/9}}$ diverges by the $p$-test. Therefore, by the Comparison Test, the original series $\sum \frac{2}{\sqrt[9]{n^2+9}}$ must diverge.

So, while the Limit Comparison Test can be applied to show convergence, it is not the best choice here, as it would give an incorrect answer. The best choice is A, diverges.

Use the Limit Comparison Test (If Possible) to Determine Whether ∑n=1∞2n2+99\sum _ { n = 1 } ^ { \infty } \frac { 2 } { \sqrt [ 9 ] { n ^ { 2 } + 9 } }n=1∑∞​9n2+9​2​

Use the Limit Comparison Test (If Possible) to Determine Whether $\sum _ { n = 1 } ^ { \infty } \frac { 2 } { \sqrt [ 9 ] { n ^ { 2 } + 9 } }$