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4. \( \displaystyle \sum_{n=1}^\infty \frac{n!}{10^n} \).

This diverges by the divergence test: the terms don't go to 0. In fact they go to \(\infty\).

8. \( \displaystyle \sum_{n=1}^\infty \frac{(-2)^n}{3^n} \).

Since \( (-2)^n/3^n = (-2/3)^n \) this is is a geometrice series. Since \( |-2/3| \lt 1 \), it converges. In fact, it converges to \( r/(1-r) \) with \( r = -2/3 \).

16. \( \displaystyle \sum_{n=1}^\infty \frac{n\ln n}{2^n} \).

Use the ratio test: \[ \frac{a_{n+1}}{a_n} = \frac{(n+1)\ln(n+1)}{n\ln n} \frac{2^n}{2^{n+1}} = \frac{n+1}{n} \frac{\ln(n+1)}{\ln n} \frac{1}{2} \] Clearly \( (n+1)/n \) goes to 1. Using l'Hospital's rule you can show that \( \ln(n+1)/\ln n \) also goes to 1. So \( a_{n+1}/a_n \to 1/2 \). So the series converges.

28. For this one \[ \frac{a_{n+1}}{a_n} = \frac{1+\arctan n}{n} \to 0, \] since \( \arctan n \le \pi/2 \). So the series converges.

42. \( \displaystyle \sum_{n=1}^\infty \frac{n^n}{(2^n)^2} \).

Notice \( (2^n)^2 = 4^n \). So the terms don't go to 0, so it diverges.

8. \( \displaystyle \sum_{n=1}^\infty (-1)^n \ln\bigg(1+\frac{1}{n}\bigg ) \).

This converges by the alternating series test.

16. \( \displaystyle \sum_{n=1}^\infty (-1)^{n+1} \frac{n!}{2^n}\).

This diverges since the terms don't go to 0.

22. \( \displaystyle \sum_{n=1}^\infty \frac{(-2)^{n+1}}{n+5^n} \)

This converges absolutely. You can see this using the comparison test (with \(2 \sum_{n=1}^\infty (2/5)^n \) or the ratio or root test.

42. \( \displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n + \sqrt{n+1}} \)

This converges by the alternating series test but it does not converge absolutely since \[ \sum_{n=1}^\infty \frac{1}{\sqrt n + \sqrt{n+1}} \ge \sum_{n=1}^\infty \frac{1}{2 \sqrt{n+1}} \] which diverges by the \(p\)-test.

46. This is an alternating series and the first unused term is \( 1/10^5 \) so that bounds the error.