(5 points)\ntest each of the following series for convergence by either the comparison test or the limit…

Explanation:

Step1: Analyze series $\sum_{n = 1}^{\infty}\frac{6n^{4}}{n^{6}+3}$

For large $n$, the dominant terms in the numerator and denominator are considered. The series behaves like $\sum_{n = 1}^{\infty}\frac{6n^{4}}{n^{6}}=\sum_{n = 1}^{\infty}\frac{6}{n^{2}}$. By the $p -$ series test ($\sum_{n = 1}^{\infty}\frac{1}{n^{p}}$ converges for $p>1$), and using the limit - comparison test with $a_{n}=\frac{6n^{4}}{n^{6}+3}$ and $b_{n}=\frac{6}{n^{2}}$. Calculate $\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\lim_{n\rightarrow\infty}\frac{\frac{6n^{4}}{n^{6}+3}}{\frac{6}{n^{2}}}=\lim_{n\rightarrow\infty}\frac{n^{6}}{n^{6}+3}=1$. Since $\sum_{n = 1}^{\infty}b_{n}$ converges, $\sum_{n = 1}^{\infty}a_{n}$ converges.

Step2: Analyze series $\sum_{n = 1}^{\infty}\frac{(- 1)^{n}}{8n}$

This is an alternating series of the form $\sum_{n = 1}^{\infty}(-1)^{n}a_{n}$ where $a_{n}=\frac{1}{8n}$. By the alternating - series test, $a_{n}=\frac{1}{8n}>0$, $a_{n + 1}=\frac{1}{8(n + 1)}<a_{n}$ for all $n\geq1$ and $\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\frac{1}{8n}=0$. But we need to use comparison tests. Consider the absolute - value series $\sum_{n = 1}^{\infty}\left|\frac{(-1)^{n}}{8n}\right|=\sum_{n = 1}^{\infty}\frac{1}{8n}$. This is a harmonic - like series ($\sum_{n = 1}^{\infty}\frac{1}{n}$ diverges), and by the $p -$ series test ($p = 1$), the absolute - value series diverges. The comparison tests are not applicable here, so we enter NA.

Step3: Analyze series $\sum_{n = 1}^{\infty}\frac{\cos^{3}(n)\sqrt{n}}{n^{3}}$

We know that $|\cos^{3}(n)|\leq1$. Then $\left|\frac{\cos^{3}(n)\sqrt{n}}{n^{3}}\right|\leq\frac{\sqrt{n}}{n^{3}}=\frac{1}{n^{\frac{5}{2}}}$. By the $p -$ series test ($p=\frac{5}{2}>1$), and using the comparison test, $\sum_{n = 1}^{\infty}\left|\frac{\cos^{3}(n)\sqrt{n}}{n^{3}}\right|$ converges, so $\sum_{n = 1}^{\infty}\frac{\cos^{3}(n)\sqrt{n}}{n^{3}}$ converges.

Step4: Analyze series $\sum_{n = 1}^{\infty}\frac{6n^{5}}{n^{7}+3}$

For large $n$, the series behaves like $\sum_{n = 1}^{\infty}\frac{6n^{5}}{n^{7}}=\sum_{n = 1}^{\infty}\frac{6}{n^{2}}$. Using the limit - comparison test with $a_{n}=\frac{6n^{5}}{n^{7}+3}$ and $b_{n}=\frac{6}{n^{2}}$. Calculate $\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\lim_{n\rightarrow\infty}\frac{\frac{6n^{5}}{n^{7}+3}}{\frac{6}{n^{2}}}=\lim_{n\rightarrow\infty}\frac{n^{7}}{n^{7}+3}=1$. Since $\sum_{n = 1}^{\infty}b_{n}$ converges, $\sum_{n = 1}^{\infty}a_{n}$ converges.

Step5: Analyze series $\sum_{n = 1}^{\infty}\frac{\cos(n)\sqrt{n}}{6n + 3}$

We know that $|\cos(n)|\leq1$, so $\left|\frac{\cos(n)\sqrt{n}}{6n + 3}\right|\leq\frac{\sqrt{n}}{6n + 3}$. For large $n$, $\frac{\sqrt{n}}{6n+3}\sim\frac{\sqrt{n}}{6n}=\frac{1}{6\sqrt{n}}$. By the $p -$ series test ($p=\frac{1}{2}<1$), the series $\sum_{n = 1}^{\infty}\frac{1}{6\sqrt{n}}$ diverges. Using the limit - comparison test with $a_{n}=\frac{\cos(n)\sqrt{n}}{6n + 3}$ and $b_{n}=\frac{1}{6\sqrt{n}}$, $\lim_{n\rightarrow\infty}\frac{\left|\frac{\cos(n)\sqrt{n}}{6n + 3}\right|}{\frac{1}{6\sqrt{n}}}=\lim_{n\rightarrow\infty}\frac{|\cos(n)|n}{6n + 3}$. Since $0\leq|\cos(n)|\leq1$, $\lim_{n\rightarrow\infty}\frac{|\cos(n)|n}{6n + 3}$ is non - zero (bounded between $0$ and $\frac{1}{6}$ for non - zero $|\cos(n)|$ values). The series diverges.

Answer:

1. CONV
2. NA
3. CONV
4. CONV
5. DIV