question 9 of 13 let $a_{n}=\frac{n^{2}}{n^{4}-5}$ and $b_{n}=\frac{1}{n^{2}}$. calculate the following limit. $lim_{n o infty}\frac{a_{n}}{b_{n}}$ (give an exact answer. use symbolic notation and fractions where needed. enter dne if the limit does not exist.) $lim_{n o infty}\frac{a_{n}}{b_{n}}=$ determine the convergence of $sum_{n = 1}^{infty}a_{n}$. $sum_{n = 1}^{infty}a_{n}$ diverges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ diverges and $lim_{n o infty}\frac{a_{n}}{b_{n}}$ is infinite. $sum_{n = 1}^{infty}a_{n}$ converges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ converges and $lim_{n o infty}\frac{a_{n}}{b_{n}}$ exists and is finite. $sum_{n = 1}^{infty}a_{n}$ converges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ converges and $lim_{n o infty}\frac{a_{n}}{b_{n}}$ does not exist. $sum_{n = 1}^{infty}a_{n}$ diverges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ diverges.

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question 9 of 13 > let $a_{n}=\frac{n^{2}}{n^{4}-5}$ and $b_{n}=\frac{1}{n^{2}}$. calculate the following limit. $lim_{n 	o infty}\frac{a_{n}}{b_{n}}$ (give an exact answer. use symbolic notation and fractions where needed. enter dne if the limit does not exist.) $lim_{n 	o infty}\frac{a_{n}}{b_{n}}=$ determine the convergence of $sum_{n = 1}^{infty}a_{n}$. $sum_{n = 1}^{infty}a_{n}$ diverges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ diverges and $lim_{n 	o infty}\frac{a_{n}}{b_{n}}$ is infinite. $sum_{n = 1}^{infty}a_{n}$ converges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ converges and $lim_{n 	o infty}\frac{a_{n}}{b_{n}}$ exists and is finite. $sum_{n = 1}^{infty}a_{n}$ converges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ converges and $lim_{n 	o infty}\frac{a_{n}}{b_{n}}$ does not exist. $sum_{n = 1}^{infty}a_{n}$ diverges by the limit comparison test because $sum_{n = 1}^{infty}b_{n}$ diverges.

Accepted Answer

# Explanation:
## Step1: Substitute $a_n$ and $b_n$ into the limit.
We have $\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=\lim_{n
ightarrow\infty}\frac{\frac{n^2}{n^4 - 5}}{\frac{1}{n^2}}$. When dividing by a fraction, we multiply by its reciprocal, so $\lim_{n
ightarrow\infty}\frac{n^2}{n^4 - 5}\times n^2=\lim_{n
ightarrow\infty}\frac{n^4}{n^4 - 5}$.
## Step2: Divide numerator and denominator by $n^4$.
$\lim_{n
ightarrow\infty}\frac{n^4}{n^4 - 5}=\lim_{n
ightarrow\infty}\frac{\frac{n^4}{n^4}}{\frac{n^4}{n^4}-\frac{5}{n^4}}=\lim_{n
ightarrow\infty}\frac{1}{1 - \frac{5}{n^4}}$.
## Step3: Evaluate the limit as $n
ightarrow\infty$.
As $n
ightarrow\infty$, $\frac{5}{n^4}
ightarrow0$. So $\lim_{n
ightarrow\infty}\frac{1}{1 - \frac{5}{n^4}}=\frac{1}{1-0}=1$.

For the convergence of $\sum_{n = 1}^{\infty}a_n$, we know that the $p -$ series $\sum_{n=1}^{\infty}b_n=\sum_{n = 1}^{\infty}\frac{1}{n^2}$ is a $p -$ series with $p = 2>1$, so $\sum_{n=1}^{\infty}b_n$ converges. Since $\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=1$ (a positive - finite number), by the Limit Comparison Test, $\sum_{n = 1}^{\infty}a_n$ converges.

# Answer:
$\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=1$
$\sum_{n = 1}^{\infty}a_n$ converges by the Limit Comparison Test because $\sum_{n = 1}^{\infty}b_n$ converges and $\lim_{n
ightarrow\infty}\frac{a_n}{b_n}$ exists and is finite.

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Question

Explanation:

Step1: Substitute \(a_n\) and \(b_n\) into the limit.

Step2: Divide numerator and denominator by \(n^4\).

Step3: Evaluate the limit as \(n

Answer: