Use the limit comparison test to determine if $$ \sum_{n=2}^{\infty} \frac{5 n^{2}+2}{3 n^{4}-5 n^{3}-6} $$ converges or diverges, and justify your answer. Answer Attempt out of 2 Apply the comparison test with the series $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ 2. If $$ a_{n}=\frac{5 n^{2}+2}{3 n^{4}-5 n^{3}-6} $$ and $$ b_{n}=\frac{1}{n^{2}} $$, 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. $$ \sum_{n=2}^{\infty} \frac{1}{n^{2}} $$ converges since a p-series will converge if and only if $$ \mathrm{p}1 \backsim $$. Therefore, $$ \sum_{n=2}^{\infty} \frac{5 n^{2}+2}{3 n^{4}-5 n^{3}-6} $$ converges

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Use the limit comparison test to determine if $$ \sum_{n=2}^{\infty} \frac{5 n^{2}+2}{3 n^{4}-5 n^{3}-6} $$ converges or diverges, and justify your answer. Answer Attempt out of 2 Apply the comparison test with the series $$ \sum_{n=2}^{\infty} \frac{1}{n^{p}} $$ where $$ p= $$ 2. If $$ a_{n}=\frac{5 n^{2}+2}{3 n^{4}-5 n^{3}-6} $$ and $$ b_{n}=\frac{1}{n^{2}} $$, 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. $$ \sum_{n=2}^{\infty} \frac{1}{n^{2}} $$ converges since a p-series will converge if and only if $$ \mathrm{p}>1 \backsim $$. Therefore, $$ \sum_{n=2}^{\infty} \frac{5 n^{2}+2}{3 n^{4}-5 n^{3}-6} $$ converges

UpStudy AI · Accepted Answer

1. Identify the given series and a comparison series. We have
   $$
   a_n = \frac{5n^2+2}{3n^4-5n^3-6},
   $$
   and we compare it with the $$ p $$-series
   $$
   b_n = \frac{1}{n^2}.
   $$
   The $$ p $$-series converges because $$ p=2>1 $$.

2. Compute the limit of the ratio:
   $$
   \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{\frac{5n^2+2}{3n^4-5n^3-6}}{\frac{1}{n^2}}
   = \lim_{n\to\infty}\frac{(5n^2+2)n^2}{3n^4-5n^3-6}
   = \lim_{n\to\infty}\frac{5n^4+2n^2}{3n^4-5n^3-6}.
   $$

3. Divide the numerator and denominator by $$ n^4 $$:
   $$
   \lim_{n\to\infty}\frac{5+\frac{2}{n^2}}{3-\frac{5}{n}-\frac{6}{n^4}}.
   $$
   As $$ n \to \infty $$, the terms $$ \frac{2}{n^2} $$, $$ \frac{5}{n} $$, and $$ \frac{6}{n^4} $$ approach $$ 0 $$; hence,
   $$
   \lim_{n\to\infty}\frac{a_n}{b_n} = \frac{5}{3}.
   $$

4. Since the limit is a finite positive number, the limit comparison test implies that both series $$ \sum a_n $$ and $$ \sum b_n $$ either converge or diverge. We already know that $$ \sum_{n=2}^{\infty}\frac{1}{n^2} $$ converges.

5. Therefore, by the limit comparison test, the series
   $$
   \sum_{n=2}^{\infty}\frac{5n^{2}+2}{3n^{4}-5n^{3}-6}
   $$
   converges.
The series $$ \sum_{n=2}^{\infty} \frac{5n^{2}+2}{3n^{4}-5n^{3}-6} $$ converges because, using the limit comparison test with $$ \sum_{n=2}^{\infty} \frac{1}{n^{2}} $$, the limit of the ratio $$ \frac{a_n}{b_n} $$ is $$ \frac{5}{3} $$, which is a finite positive number. Since $$ \sum_{n=2}^{\infty} \frac{1}{n^{2}} $$ converges, the given series also converges.

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