Use the limit comparison test to determine if $$ \sum_{n=1}^{\infty} \frac{4 n^{\frac{1}{3}}}{6 n+3 n^{\frac{2}{3}}-2} $$ converges or diverges, and justify your answer. Answer Attempt 2 out of 2 Apply the comparison test with the series $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ I. If $$ a_{n}=\frac{4 n^{\frac{1}{3}}}{6 n+3 n^{\frac{2}{3}}-2} $$ 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. $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ diverges since a p-series will diverge if and only if $$ \sum_{n=1}^{\infty} \frac{4 n^{\frac{1}{3}}}{6 n+3 n^{\frac{2}{3}}-2} $$. Therefore,

Question

Use the limit comparison test to determine if $$ \sum_{n=1}^{\infty} \frac{4 n^{\frac{1}{3}}}{6 n+3 n^{\frac{2}{3}}-2} $$ converges or diverges, and justify your answer. Answer Attempt 2 out of 2 Apply the comparison test with the series $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ I. If $$ a_{n}=\frac{4 n^{\frac{1}{3}}}{6 n+3 n^{\frac{2}{3}}-2} $$ 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. $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ diverges since a p-series will diverge if and only if $$ \sum_{n=1}^{\infty} \frac{4 n^{\frac{1}{3}}}{6 n+3 n^{\frac{2}{3}}-2} $$. Therefore,

UpStudy AI · Accepted Answer

Let 
$$
a_n = \frac{4 n^{\frac{1}{3}}}{6 n + 3 n^{\frac{2}{3}} - 2}
$$
and choose the comparison series
$$
b_n = \frac{1}{n^{\frac{2}{3}}}.
$$

**Step 1.** Determine the dominant behavior of $$a_n$$ for large $$n$$.

For large $$n$$:
- The numerator is $$4n^{\frac{1}{3}}$$.
- The denominator is dominated by the term $$6n$$ (since $$6n \gg 3 n^{\frac{2}{3}}$$ and $$-2$$ as $$n \to \infty$$).

Thus, for large $$n$$,
$$
a_n \approx \frac{4n^{\frac{1}{3}}}{6n} = \frac{4}{6} n^{-\frac{2}{3}} = \frac{2}{3} \cdot \frac{1}{n^{\frac{2}{3}}}.
$$
This motivates the choice $$b_n = \frac{1}{n^{\frac{2}{3}}}$$.

**Step 2.** Compute the limit of the ratio $$\frac{a_n}{b_n}$$.

We have:
$$
\frac{a_n}{b_n} = \frac{\frac{4 n^{\frac{1}{3}}}{6 n + 3 n^{\frac{2}{3}} - 2}}{\frac{1}{n^{\frac{2}{3}}}} 
= \frac{4 n^{\frac{1}{3}} \cdot n^{\frac{2}{3}}}{6 n + 3 n^{\frac{2}{3}} - 2}
= \frac{4 n}{6 n + 3 n^{\frac{2}{3}} - 2}.
$$

Factor $$n$$ from the denominator:
$$
\frac{4 n}{n\left(6 + 3n^{-\frac{1}{3}} - \frac{2}{n}\right)} = \frac{4}{6 + 3n^{-\frac{1}{3}} - \frac{2}{n}}.
$$

Taking the limit as $$n \to \infty$$:
$$
\lim_{n\to\infty} \frac{a_n}{b_n} = \frac{4}{6 + 0 - 0} = \frac{4}{6} = \frac{2}{3}.
$$

Since the limit is a finite positive number ($$\frac{2}{3}$$), by the limit comparison test, $$\sum a_n$$ and $$\sum b_n$$ have the same convergence behavior.

**Step 3.** Analyze the convergence of the $$p$$-series $$\sum b_n$$.

The series $$\sum_{n=1}^{\infty} \frac{1}{n^{p}}$$ converges if and only if $$p > 1$$. In our case, $$p = \frac{2}{3}$$, which means
$$
\sum_{n=1}^{\infty} \frac{1}{n^{\frac{2}{3}}}
$$
diverges.

**Conclusion:**  
By the limit comparison test, the series
$$
\sum_{n=1}^{\infty} \frac{4 n^{\frac{1}{3}}}{6 n + 3 n^{\frac{2}{3}} - 2}
$$
diverges.
The series $$ \sum_{n=1}^{\infty} \frac{4 n^{\frac{1}{3}}}{6 n + 3 n^{\frac{2}{3}} - 2} $$ diverges.

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