Use the limit comparison test to determine if $$ \sum_{n=2}^{\infty} \frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ converges or diverges, and justify your answer. Answer Attempt 1out of 2 Apply the comparison test with the series $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ . If $$ a_{n}=\frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ and $$ b_{n}=\frac{1}{n^{p}} $$, then $$ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\square $$. Since $$ a_{n}, b_{n}0 $$ and the limit is a finite and positive (non-zero) number, the limit comparison test applies. $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ if and only if since a p-series will .Therefore, $$ \sum_{n=2}^{\infty} \frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$

Question

Use the limit comparison test to determine if $$ \sum_{n=2}^{\infty} \frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ converges or diverges, and justify your answer. Answer Attempt 1out of 2 Apply the comparison test with the series $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ . If $$ a_{n}=\frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$ and $$ b_{n}=\frac{1}{n^{p}} $$, then $$ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\square $$. Since $$ a_{n}, b_{n}>0 $$ and the limit is a finite and positive (non-zero) number, the limit comparison test applies. $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ if and only if since a p-series will .Therefore, $$ \sum_{n=2}^{\infty} \frac{2}{4 n^{\frac{1}{2}}-4 n^{\frac{1}{6}}-1} $$

UpStudy AI · Accepted Answer

We start by considering the series

$$
\sum_{n=2}^{\infty} \frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}.
$$

For large $$ n $$, the dominant term in the denominator $$ 4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1 $$ is $$ 4n^{\frac{1}{2}} $$ since $$ n^{\frac{1}{2}} $$ grows much faster than $$ n^{\frac{1}{6}} $$ and the constant term becomes negligible. Hence, asymptotically we have

$$
\frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1} \sim \frac{2}{4n^{\frac{1}{2}}} = \frac{1}{2n^{\frac{1}{2}}}.
$$

This suggests we compare with the $$ p $$-series

$$
\sum_{n=2}^{\infty} \frac{1}{n^p} \quad \text{with} \quad p = \frac{1}{2}.
$$

Recall that a $$ p $$-series converges if and only if $$ p > 1 $$. Since $$ \frac{1}{2} < 1 $$, the series

$$
\sum_{n=2}^{\infty} \frac{1}{n^{\frac{1}{2}}}
$$

diverges.

Next, apply the limit comparison test with

$$
a_n = \frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1} \quad \text{and} \quad b_n = \frac{1}{n^{\frac{1}{2}}}.
$$

Compute the limit:

$$
\lim_{n \to \infty} \frac{a_n}{b_n} 
= \lim_{n \to \infty} \frac{\frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}}{\frac{1}{n^{\frac{1}{2}}}}
= \lim_{n \to \infty} \frac{2n^{\frac{1}{2}}}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}.
$$

To simplify, divide the numerator and the denominator by $$ n^{\frac{1}{2}} $$:

$$
\lim_{n \to \infty} \frac{2}{4 - 4n^{-\frac{1}{3}} - n^{-\frac{1}{2}}} = \frac{2}{4 - 0 - 0} = \frac{2}{4} = \frac{1}{2}.
$$

Since the limit is positive and finite ($$ \frac{1}{2} $$), the limit comparison test applies. Therefore, $$ \sum a_n $$ has the same convergence behavior as the $$ p $$-series $$ \sum b_n $$.

Since the $$ p $$-series with $$ p = \frac{1}{2} $$ diverges, we conclude that

$$
\sum_{n=2}^{\infty} \frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1}
$$

diverges.
The series $$ \sum_{n=2}^{\infty} \frac{2}{4n^{\frac{1}{2}} - 4n^{\frac{1}{6}} - 1} $$ diverges because it behaves similarly to the divergent $$ p $$-series $$ \sum_{n=2}^{\infty} \frac{1}{n^{\frac{1}{2}}} $$ as $$ n $$ becomes large.

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