Read Theorem 5.12 in the textbook Links to an external site. . Suppose ð ð ⥠0  for all ð ⥠1 and ð ð ⥠0  for all ð ⥠1 . a) If â ð = 1 â ð ð  converges and lim ð â â ð ð ð ð = 5  then  â ð = 1 â ð ð   [ Select ] might converge or diverge (there is not enough information) converges diverges  b) If â ð = 1 â ð ð  converges and lim ð â â ð ð ð ð = 0   then  â ð = 1 â ð ð   [ Select ] diverges converges might converge or diverge (there is not enough information)  c) If â ð = 1 â ð ð  diverges and lim ð â â ð ð ð ð = 0   then  â ð = 1 â ð ð   [ Select ] might converge or diverge (there is not enough information) diverges converges  d) If â ð = 1 â ð ð  diverges and lim ð â â ð ð ð ð = 200   then  â ð = 1 â ð ð   divergeså€éäžæéæ©é¢
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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 ±â±â\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 ââ \infty NIF | | if the series equals to ââââ -\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 ââð=0ððân=0âqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ðpp-series, where ð>1p>1p > 1 LP | | Comparison with ðpp-series, where ð<1p<1p < 1 [/table] [table] âð=1â(ð+1)7(8â ð!)2(2ð)!ân=1â(n+1)7(8â n!)2(2n)! \sum_{n=1}^\infty (n+1)^{{7}} \,\, \frac{({8}\cdot n!)^2 }{ (2n)! } | | because âð=1â(1â6ð)2ð2ân=1â(1â6n)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 ±â±â\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 ââ \infty NIF | | if the series equals to ââââ -\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 ââð=0ððân=0âqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ðpp-series, where ð>1p>1p > 1 LP | | Comparison with ðpp-series, where ð<1p<1p < 1 [/table] [table] âð=1âsin(ð9ð)ân=1âsinâ¡(Ï9n) \sum_{n=1}^\infty \sin\left( \frac{\pi }{ {9} n } \right) | | because âð=1âcos(ð10ð)ân=1âcosâ¡(Ï10n) \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 ±â±â\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 ââ \infty NIF | | if the series equals to ââââ -\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 ââð=0ððân=0âqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ðpp-series, where ð>1p>1p > 1 LP | | Comparison with ðpp-series, where ð<1p<1p < 1 [/table] [table] âð=1â3ð3+15ð12+1âŸâŸâŸâŸâŸâŸâŸâ3+ðân=1â3n3+15n12+13+n \sum_{n=1}^\infty \frac{{3}n^{3}+1}{{5}\sqrt[{3}]{ n^{12}+1}+n} | | because âð=1â(ðlnð)24ðân=1â(nlnâ¡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 â{0,1,tan(1/6)}â{0,1,tanâ¡(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)tanâ¡(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 ââð=0ððân=0âqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ðpp-series, where ð>1p>1p > 1 LP | | Comparison with ðpp-series, where ð<1p<1p < 1 [/table] [table] âð=1â(â1)ðtan(16ð)ân=1â(â1)ntanâ¡(16n) \sum\limits_{n=1}^{\infty} (-1)^n \tan \left(\frac{1}{{6} n}\right) | | because âð=1â(â1)ð(1â7ð)ðân=1â(â1)n(1â7n)n\sum\limits_{n=1}^{\infty} (-1)^n \left(1-\frac{{7}}{n}\right)^n | | because [/table]
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