Question

let (a_{n}=\frac{5n^{2}+14n}{3n^{4}-5n^{2}-22}), (b_{n}=\frac{5}{3n^{2}}). calculate the limit. (give an exact answer. use symbolic notation and fractions where needed. enter dne if the limit does not exist.) (lim_{n
ightarrowinfty}\frac{a_{n}}{b_{n}}=) determine the convergence or divergence of (sum_{n = 1}^{infty}a_{n}). (sum_{n = 1}^{infty}a_{n}) diverges by the limit comparison test because (lim_{n
ightarrowinfty}\frac{a_{n}}{b_{n}}) is finite and (sum_{n = 1}^{infty}b_{n}) diverges. (sum_{n = 1}^{infty}a_{n}) converges by the limit comparison test because (lim_{n
ightarrowinfty}\frac{a_{n}}{b_{n}}) is finite and (sum_{n = 1}^{infty}b_{n}) diverges. (sum_{n = 1}^{infty}a_{n}) converges by the limit comparison test because (lim_{n
ightarrowinfty}\frac{a_{n}}{b_{n}}) is finite and (sum_{n = 1}^{infty}b_{n}) converges. it is not possible to use the limit comparison test to determine the convergence or divergence of (sum_{n = 1}^{infty}a_{n}).

Explanation:

Step1: Write the ratio of \(a_n\) and \(b_n\)

$$ \frac{a_n}{b_n}=\frac{\frac{5n^2 + 14n}{3n^4-5n^2 - 22}}{\frac{5}{3n^2}}=\frac{(5n^2 + 14n)\times3n^2}{5\times(3n^4-5n^2 - 22)} $$

Step2: Expand the numerator and denominator

The numerator is \((5n^2 + 14n)\times3n^2=15n^4+42n^3\). The denominator is \(15n^4-25n^2 - 110\).

Step3: Divide both numerator and denominator by \(n^4\)

$$ \lim_{n ightarrow\infty}\frac{a_n}{b_n}=\lim_{n ightarrow\infty}\frac{15 + \frac{42}{n}}{15-\frac{25}{n^2}-\frac{110}{n^4}} $$

Step4: Evaluate the limit

As \(n
ightarrow\infty\), \(\frac{42}{n}
ightarrow0\), \(\frac{25}{n^2}
ightarrow0\) and \(\frac{110}{n^4}
ightarrow0\). So \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=\frac{15 + 0}{15-0 - 0}=1\)

For the convergence - divergence part:
The series \(\sum_{n = 1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{5}{3n^2}\) is a \(p\) - series with \(p = 2>1\), so \(\sum_{n = 1}^{\infty}b_n\) converges.
Since \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}=1\) (finite) and \(\sum_{n = 1}^{\infty}b_n\) converges, 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 \(\lim_{n
ightarrow\infty}\frac{a_n}{b_n}\) is finite and \(\sum_{n=1}^{\infty}b_n\) converges.