Read Theorem 5.12 in the textbook Links to an external site. .  Suppose 𝑎 𝑛 ≥ 0   for all 𝑛 ≥ 1 and 𝑏 𝑛 ≥ 0   for all 𝑛 ≥ 1 . a) If ∑ 𝑛 = 1 ∞ 𝑏 𝑛   converges and lim 𝑛 → ∞ 𝑎 𝑛 𝑏 𝑛 = 5   then   ∑ 𝑛 = 1 ∞ 𝑎 𝑛     [ Select ] might converge or diverge (there is not enough information) converges diverges   b) If ∑ 𝑛 = 1 ∞ 𝑏 𝑛   converges and lim 𝑛 → ∞ 𝑎 𝑛 𝑏 𝑛 = 0     then   ∑ 𝑛 = 1 ∞ 𝑎 𝑛     [ Select ] diverges converges might converge or diverge (there is not enough information)   c) If ∑ 𝑛 = 1 ∞ 𝑏 𝑛   diverges and lim 𝑛 → ∞ 𝑎 𝑛 𝑏 𝑛 = 0     then   ∑ 𝑛 = 1 ∞ 𝑎 𝑛     [ Select ] might converge or diverge (there is not enough information) diverges converges   d) If ∑ 𝑛 = 1 ∞ 𝑏 𝑛   diverges and lim 𝑛 → ∞ 𝑎 𝑛 𝑏 𝑛 = 200     then   ∑ 𝑛 = 1 ∞ 𝑎 𝑛     diverges倚重䞋拉选择题

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Question textFor each of the series below you find two answer fields. In the first answer field enter: (inputs are case sensitive) [table] DV | | if the series is divergent and not equal to ±∞±∞\pm\infty CV | | if the series is convergent (to a non-zero number) Z | | if the series converges to 0 INF | | if the series equals to ∞∞ \infty NIF | | if the series equals to −∞−∞ -\infty [/table] [table] WD | | if the series is not well defined | | [/table] In the second answer field select one of the following reasons that can be used to prove your claim in the first answer field: [table] DT | | The Divergency Test IT | | The Integral Test AS | | The Alternating Series Test RO | | The Root Test RA | | The Ratio Test [/table] [table] D | | The sequence of summands decreases to 00 0 L | | The limit of summands exists and equals to 00 0 C | | Comparison with a  geometric series ∑∞𝑛=0𝑞𝑛∑n=0∞qn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with 𝑝pp-series, where 𝑝>1p>1p > 1 LP | | Comparison with 𝑝pp-series, where 𝑝<1p<1p < 1 [/table] [table] ∑𝑛=1∞(𝑛+1)7(8⋅𝑛!)2(2𝑛)!∑n=1∞(n+1)7(8⋅n!)2(2n)! \sum_{n=1}^\infty (n+1)^{{7}} \,\, \frac{({8}\cdot n!)^2 }{ (2n)! } | | because ∑𝑛=1∞(1−6𝑛)2𝑛2∑n=1∞(1−6n)2n2 \sum\limits_{n=1}^{\infty} \left(1- \frac{{6} }{ n } \right)^{ {2}n^2} | | because [/table]

Question textFor each of the series below you find two answer fields. In the first answer field enter: (inputs are case sensitive) [table] DV | | if the series is divergent and not equal to ±∞±∞\pm\infty CV | | if the series is convergent (to a non-zero number) Z | | if the series converges to 0 INF | | if the series equals to ∞∞ \infty NIF | | if the series equals to −∞−∞ -\infty [/table] [table] WD | | if the series is not well defined | | [/table] In the second answer field select one of the following reasons that can be used to prove your claim in the first answer field: [table] DT | | The Divergency Test IT | | The Integral Test AS | | The Alternating Series Test RO | | The Root Test RA | | The Ratio Test [/table] [table] D | | The sequence of summands decreases to 00 0 L | | The limit of summands exists and equals to 00 0 C | | Comparison with a  geometric series ∑∞𝑛=0𝑞𝑛∑n=0∞qn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with 𝑝pp-series, where 𝑝>1p>1p > 1 LP | | Comparison with 𝑝pp-series, where 𝑝<1p<1p < 1 [/table] [table] ∑𝑛=1∞sin(𝜋9𝑛)∑n=1∞sin⁡(π9n) \sum_{n=1}^\infty \sin\left( \frac{\pi }{ {9} n } \right) | | because ∑𝑛=1∞cos(𝜋10𝑛)∑n=1∞cos⁡(π10n) \sum\limits_{n=1}^{\infty} \cos\left( \frac{\pi }{ {10} n } \right) | | because [/table]

Question textFor each of the series below you find two answer fields. In the first answer field enter: (inputs are case sensitive) [table] DV | | if the series is divergent and not equal to ±∞±∞\pm\infty CV | | if the series is convergent (to a non-zero number) Z | | if the series converges to 0 INF | | if the series equals to ∞∞ \infty NIF | | if the series equals to −∞−∞ -\infty [/table] [table] WD | | if the series is not well defined | | [/table] In the second answer field select one of the following reasons that can be used to prove your claim in the first answer field: [table] DT | | The Divergency Test IT | | The Integral Test AS | | The Alternating Series Test RO | | The Root Test RA | | The Ratio Test [/table] [table] D | | The sequence of summands decreases to 00 0 L | | The limit of summands exists and equals to 00 0 C | | Comparison with a  geometric series ∑∞𝑛=0𝑞𝑛∑n=0∞qn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with 𝑝pp-series, where 𝑝>1p>1p > 1 LP | | Comparison with 𝑝pp-series, where 𝑝<1p<1p < 1 [/table] [table] ∑𝑛=1∞3𝑛3+15𝑛12+1‟‟‟‟‟‟‟√3+𝑛∑n=1∞3n3+15n12+13+n \sum_{n=1}^\infty \frac{{3}n^{3}+1}{{5}\sqrt[{3}]{ n^{12}+1}+n} | | because ∑𝑛=1∞(𝑛ln𝑛)24𝑛∑n=1∞(nln⁡n)24n \sum\limits_{n=1}^{\infty} \frac{(n \ln n)^{2}}{{4}^n} | | because [/table]

Question textFor each of the series below you find two answer fields. In the first answer field enter: (inputs are case sensitive) [table] DV | | if the series is divergent CV | | if the series is convergent (to a number ∉{0,1,tan(1/6)}∉{0,1,tan⁡(1/6)}\not\in \{0, 1, \tan(1/{6}) \}) Z | | if the series converges to 0 ON | | if the series converges to 111 IN | | if the series equals to tan(1/6)tan⁡(1/6)\tan(1/{6}) [/table] [table] WD | | if the series is not well defined | | [/table] In the second answer field select one of the following reasons that can be used to prove your claim in the first answer field: [table] DT | | The Divergency Test IT | | The Integral Test AST | | The Alternating Series Test RO | | The Root Test RA | | The Ratio Test [/table] [table] D | | The sequence of summands decreases to 00 0 L | | The limit of summands exists and equals to 00 0 C | | Comparison with a  geometric series ∑∞𝑛=0𝑞𝑛∑n=0∞qn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with 𝑝pp-series, where 𝑝>1p>1p > 1 LP | | Comparison with 𝑝pp-series, where 𝑝<1p<1p < 1 [/table] [table] ∑𝑛=1∞(−1)𝑛tan(16𝑛)∑n=1∞(−1)ntan⁡(16n) \sum\limits_{n=1}^{\infty} (-1)^n \tan \left(\frac{1}{{6} n}\right) | | because ∑𝑛=1∞(−1)𝑛(1−7𝑛)𝑛∑n=1∞(−1)n(1−7n)n\sum\limits_{n=1}^{\infty} (-1)^n \left(1-\frac{{7}}{n}\right)^n | | because [/table]

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