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Section 9.5 #7: Determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n}{3^n}.$
Solution: Compute $$\displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{3^n}=0.$$ Since this series is alternating, the alternating series test allows us to conclude that it converges.

Section 9.5 #9: Determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n(5n-1)}{4n+1}.$
Solution: This series diverges because $\displaystyle\lim_{n \rightarrow \infty} \dfrac{(-1)^n(5n-1)}{4n+1}$ does not exist (it must be zero for the series to possibly converge). This limit does not exist because the limit of $\dfrac{5n-1}{4n+1}$ is $\dfrac{5}{4}$ while the factor $(-1)^n$ oscillates forever.

Section 9.6 #14: Use the ratio test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n!}$.
Solution: Apply the ratio test by computing $$\displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{1}{(n+1)!}}{\frac{1}{n!}} = \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{n+1} = 0,$$ and so by the ratio test, this series converges.

Section 9.6 #15: Use the ratio test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{3^n}$.
Solution: Apply the ratio test by computing $$\displaystyle\lim_{n \rightarrow \infty} \dfrac{\frac{(n+1)!}{3^{n+1}}}{\frac{n!}{3^n}} = \displaystyle\lim_{n \rightarrow \infty} \dfrac{n+1}{3} = \infty,$$ and so we may conclude that the series diverges.

Section 9.6 #32: Use the ratio test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n 2^{4n}}{(2n+1)!}$.
Solution: Apply the ratio test by computing $$\displaystyle\lim_{n \rightarrow \infty} \left| \dfrac{\frac{2^{4(n+1)}}{(2(n+1)+1)!}}{\frac{2^{4n}}{(2n+1)!}} \right| = \displaystyle\lim_{n \rightarrow \infty} \left| \dfrac{2^4}{(2n+3)(2n+2)} \right|=0,$$ and so we may conclude that the series converges.

Section 9.6 #36: Use the root test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^n}$.
Solution: Apply the root test by computing $$\displaystyle\lim_{n \rightarrow \infty} \sqrt[n]{\dfrac{1}{n^n}} = \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{n} = 0,$$ and so by the root test, we may conclude that the series converges.

Section 9.6 #37: Use the root test to determine the convergence or divergence of the series: $\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{n}{2n+1} \right)^n$.
Solution: Apply the root test by computing $$\displaystyle\lim_{n \rightarrow \infty} \sqrt[n]{ \left( \dfrac{n}{2n+1} \right)^n} = \displaystyle\lim_{n \rightarrow \infty} \dfrac{n}{2n+1} = \dfrac{1}{2},$$ and so the root test allows us to conclude that the series converges.

Section 9.6 #54: Determine the convergence or divergence of the series $\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{2\pi}{3} \right)^n$.
Solution: The series diverges because $\dfrac{2\pi}{3} \approx 2.09 \gt 1$ (the geometric series diverges if the ratio is bigger than $1$!).

Section 9.6 #62: Determine the convergence or divergence of the series $\displaystyle\sum_{n=1}^{\infty} \left( \dfrac{(-1)^n}{n \ln(n)} \right)^n$.
Solution: Applying the root test, $$\displaystyle\lim_{n \rightarrow \infty} \sqrt[n]{ \left| \left( \dfrac{(-1)^n}{n\ln(n)} \right)^n \right| } = \displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{n\ln(n)} = 0,$$ and so by the root test we may conclude that the series converges.