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HW 9Sec 9.54. \( \displaystyle \sum_{n=1}^\infty \frac{n!}{10^n} \). This diverges by the divergence test: the terms don't go to 0. In fact they go to \(\infty\).
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 \( r = -2/3 \).
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.
Notice \( (2^n)^2 = 4^n \). So the terms don't go to 0, so it diverges. Sec 9.6
This converges by the alternating series test.
This diverges since the terms don't go to 0.
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.
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.
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