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Section 9.5 #7: Determine the convergence or divergence of the series: n=1(1)n3n.
Solution: Compute lim 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.