Use the limit comparison test to determine if $$ \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6 n+2} $$ 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 $$ p= $$ $$ \square $$ . If $$ a_{n}=\frac{3 n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6 n+2} $$ 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. $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ since a p-series will $$ \qquad $$ if and only if $$ \square $$ . Therefore, $$ \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6 n+2} $$

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Use the limit comparison test to determine if $$ \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6 n+2} $$ 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 $$ p= $$ $$ \square $$ . If $$ a_{n}=\frac{3 n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6 n+2} $$ 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. $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ since a p-series will $$ \qquad $$ if and only if $$ \square $$ . Therefore, $$ \sum_{n=1}^{\infty} \frac{3 n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6 n+2} $$

UpStudy AI · Accepted Answer

1. For large $$ n $$, the dominant term in the numerator is $$ 3n^{\frac{1}{6}} $$ and in the denominator is $$ n^{\frac{5}{3}} $$. Thus,  
   $$
   a_n=\frac{3n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6n+2} \sim \frac{3n^{\frac{1}{6}}}{n^{\frac{5}{3}}} = \frac{3}{n^{\frac{5}{3}-\frac{1}{6}}} = \frac{3}{n^{\frac{9}{6}}} = \frac{3}{n^{\frac{3}{2}}}.
   $$

2. Choose the comparison series  
   $$
   b_n=\frac{1}{n^{p}}
   $$
   with  
   $$
   p=\frac{3}{2}.
   $$
   The $$ p $$-series $$\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$$ converges since $$ \frac{3}{2}>1 $$.

3. Compute the limit comparison:
   $$
   \lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{3n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6n+2}}{\frac{1}{n^{\frac{3}{2}}}}=\lim_{n\to\infty}\frac{(3n^{\frac{1}{6}}+5)n^{\frac{3}{2}}}{n^{\frac{5}{3}}+6n+2}.
   $$
   For large $$ n $$, the dominant behavior gives:
   $$
   \frac{(3n^{\frac{1}{6}})n^{\frac{3}{2}}}{n^{\frac{5}{3}}}=\frac{3n^{\frac{1}{6}+\frac{3}{2}}}{n^{\frac{5}{3}}}=\frac{3n^{\frac{10}{6}}}{n^{\frac{10}{6}}}=3.
   $$
   Hence, 
   $$
   \lim_{n\to\infty}\frac{a_n}{b_n}=3.
   $$

4. Since this limit is a positive finite number and the $$ p $$-series $$\sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$$ converges, by the limit comparison test, the series  
   $$
   \sum_{n=1}^{\infty} \frac{3n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6n+2}
   $$
   also converges.
The series $$ \sum_{n=1}^{\infty} \frac{3n^{\frac{1}{6}}+5}{n^{\frac{5}{3}}+6n+2} $$ converges by the limit comparison test with the convergent $$ p $$-series $$ \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}} $$.

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