Use the limit comparison test to determine if $$ \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}}{3 n-3 n^{\frac{1}{6}}+2} $$ converges or diverges, and justify your answer. Answer Attempt 1 out of 2 Apply the comparison test with the series $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ $$ \frac{5}{6} $$. If $$ a_{n}=\frac{3 n^{\frac{1}{6}}}{3 n-3 n^{\frac{1}{6}}+2} $$ and $$ b_{n}=\frac{1}{n^{\frac{5}{6}}} $$, then $$ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}= $$ II Since $$ a_{n}, b_{n}0 $$ and the limit is a finite and positive (non-zero) number, the limit comparison test applies. $$ \sum_{n=1}^{\infty} \frac{1}{n^{\frac{5}{6}}} $$ diverges since a p-series will diverge if and only if $$ \mathrm{p}

Question

Use the limit comparison test to determine if $$ \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}}{3 n-3 n^{\frac{1}{6}}+2} $$ converges or diverges, and justify your answer. Answer Attempt 1 out of 2 Apply the comparison test with the series $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ $$ \frac{5}{6} $$. If $$ a_{n}=\frac{3 n^{\frac{1}{6}}}{3 n-3 n^{\frac{1}{6}}+2} $$ and $$ b_{n}=\frac{1}{n^{\frac{5}{6}}} $$, then $$ \lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}= $$ II Since $$ a_{n}, b_{n}>0 $$ and the limit is a finite and positive (non-zero) number, the limit comparison test applies. $$ \sum_{n=1}^{\infty} \frac{1}{n^{\frac{5}{6}}} $$ diverges since a p-series will diverge if and only if $$ \mathrm{p}<1 $$ $$ \infty $$

UpStudy AI · Accepted Answer

**Step 1. Identify the general term and choose a comparison series**  
We have  
$$
a_n = \frac{3 n^{\frac{1}{6}}}{3 n - 3 n^{\frac{1}{6}}+2}.
$$  
For large $$ n $$, the dominant term in the denominator is $$ 3n $$. Thus,  
$$
a_n \sim \frac{3 n^{\frac{1}{6}}}{3n} = \frac{n^{\frac{1}{6}}}{n} = \frac{1}{n^{\frac{5}{6}}}.
$$  
So we choose the comparison series  
$$
b_n = \frac{1}{n^{\frac{5}{6}}}.
$$

**Step 2. Check the convergence of the comparison series**  
The series  
$$
\sum_{n=1}^{\infty} b_n = \sum_{n=1}^{\infty} \frac{1}{n^{\frac{5}{6}}}
$$  
is a $$ p $$-series with $$ p=\frac{5}{6} $$. Since $$ p < 1 $$, the $$ p $$-series diverges.

**Step 3. Compute the limit for the Limit Comparison Test**  
We have  
$$
\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{3 n^{\frac{1}{6}}}{3 n-3 n^{\frac{1}{6}}+2}}{\frac{1}{n^{\frac{5}{6}}}} 
= \lim_{n \to \infty} \frac{3 n^{\frac{1}{6}} \cdot n^{\frac{5}{6}}}{3 n-3 n^{\frac{1}{6}}+2}.
$$  
Since $$ n^{\frac{1}{6}} \cdot n^{\frac{5}{6}} = n $$, this becomes  
$$
\lim_{n \to \infty} \frac{3 n}{3 n-3 n^{\frac{1}{6}}+2}.
$$

Divide numerator and denominator by $$ n $$:
$$
\lim_{n \to \infty} \frac{3}{3-\frac{3}{n^{\frac{5}{6}}}+\frac{2}{n}}.
$$

As $$ n \to \infty $$, both $$\frac{3}{n^{\frac{5}{6}}}$$ and $$\frac{2}{n}$$ approach 0. Thus,  
$$
\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{3}{3-0+0} = 1.
$$

**Step 4. Apply the Limit Comparison Test**  
Since  
- $$\lim_{n \to \infty} \frac{a_n}{b_n} = 1$$ (which is finite and positive), and  
- $$\sum_{n=1}^{\infty} b_n$$ diverges,

the Limit Comparison Test tells us that the series  
$$
\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}}{3 n-3 n^{\frac{1}{6}}+2}
$$  
also diverges.

**Conclusion:**  
By the Limit Comparison Test, the series diverges.
The series $$ \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}}{3 n-3 n^{\frac{1}{6}}+2} $$ diverges.

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Calculus Questions