QuestionTest each of the following series for convergence by either the Comparison Test or the Limit Comparison Test. If at least one test can be applied to the series, enter CONV if it converges or DIV if it diverges. If neither test can be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the comparison tests cannot be applied to it, then you must enter NA rather than CONV.)
1. $\sum_{n=1}^{\infty} \frac{\cos ^{2}(n) \sqrt{n}}{n^{4}}$
2. $\sum_{n=1}^{\infty} \frac{7 n^{4}}{n^{5}+7}$
3. $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{7 n}$
4. $\sum_{n=1}^{\infty} \frac{\cos (n) \sqrt{n}}{7 n+7}$
5. $\sum_{n=1}^{\infty} \frac{8 n^{7}-n^{5}+7 \sqrt{n}}{7 n^{9}-n^{4}+2}$

Studdy Solution
Consider the series $\sum_{n=1}^{\infty} \frac{8 n^{7}-n^{5}+7 \sqrt{n}}{7 n^{9}-n^{4}+2}$.
The term $\frac{8 n^{7}-n^{5}+7 \sqrt{n}}{7 n^{9}-n^{4}+2}$ is similar to $\frac{8 n^{7}}{7 n^{9}} = \frac{8}{7 n^{2}}$.
The series $\sum_{n=1}^{\infty} \frac{1}{n^{2}}$ is a $p$-series with $p = 2 > 1$, which converges.
Using the Limit Comparison Test with $b_n = \frac{1}{n^{2}}$, we compute:
$\lim_{n \to \infty} \frac{\frac{8 n^{7}-n^{5}+7 \sqrt{n}}{7 n^{9}-n^{4}+2}}{\frac{1}{n^{2}}} = \lim_{n \to \infty} \frac{8 n^{9}-n^{7}+7 n^{2.5}}{7 n^{9}-n^{4}+2} = \frac{8}{7}$
Since the limit is a positive finite number, the series converges by the Limit Comparison Test.
The results for each series are:
1. CONV
2. DIV
3. NA
4. NA
5. CONV