Math / Calculus

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

STEP 1

1. We are testing the convergence of each series using the Comparison Test or the Limit Comparison Test.
2. If neither test can be applied, we will conclude with NA.
3. We will analyze each series individually.

STEP 2

1. Analyze the first series: $\sum_{n=1}^{\infty} \frac{\cos ^{2}(n) \sqrt{n}}{n^{4}}$ .
2. Analyze the second series: $\sum_{n=1}^{\infty} \frac{7 n^{4}}{n^{5}+7}$ .
3. Analyze the third series: $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{7 n}$ .
4. Analyze the fourth series: $\sum_{n=1}^{\infty} \frac{\cos (n) \sqrt{n}}{7 n+7}$ .
5. Analyze the fifth series: $\sum_{n=1}^{\infty} \frac{8 n^{7}-n^{5}+7 \sqrt{n}}{7 n^{9}-n^{4}+2}$ .

STEP 3

Consider the series $\sum_{n=1}^{\infty} \frac{\cos ^{2}(n) \sqrt{n}}{n^{4}}$ .
The term $\frac{\cos ^{2}(n) \sqrt{n}}{n^{4}}$ is similar to $\frac{\sqrt{n}}{n^{4}} = \frac{1}{n^{3.5}}$ .
The series $\sum_{n=1}^{\infty} \frac{1}{n^{3.5}}$ is a $p$ -series with $p = 3.5 > 1$ , which converges.
Using the Comparison Test, since $0 \leq \cos^2(n) \leq 1$ , we have $0 \leq \frac{\cos ^{2}(n) \sqrt{n}}{n^{4}} \leq \frac{1}{n^{3.5}}$ .
Thus, the series converges by the Comparison Test.

STEP 4

Consider the series $\sum_{n=1}^{\infty} \frac{7 n^{4}}{n^{5}+7}$ .
The term $\frac{7 n^{4}}{n^{5}+7}$ is similar to $\frac{7 n^{4}}{n^{5}} = \frac{7}{n}$ .
The series $\sum_{n=1}^{\infty} \frac{1}{n}$ is a harmonic series, which diverges.
Using the Limit Comparison Test with $b_n = \frac{1}{n}$ , we compute:
$\lim_{n \to \infty} \frac{\frac{7 n^{4}}{n^{5}+7}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{7 n^{5}}{n^{5}+7} = 7$
Since the limit is a positive finite number, the series diverges by the Limit Comparison Test.

STEP 5

Consider the series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{7 n}$ .
This is an alternating series, and neither the Comparison Test nor the Limit Comparison Test can be applied directly to alternating series.
Thus, we conclude NA for this series.

STEP 6

Consider the series $\sum_{n=1}^{\infty} \frac{\cos (n) \sqrt{n}}{7 n+7}$ .
The term $\frac{\cos (n) \sqrt{n}}{7 n+7}$ is similar to $\frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}$ .
The series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ is a $p$ -series with $p = 0.5 < 1$ , which diverges.
Using the Comparison Test, since $-1 \leq \cos(n) \leq 1$ , we cannot establish a consistent inequality with a known convergent or divergent series.
Thus, we conclude NA for this series.

STEP 7

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

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