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MAT136H5 S 2025 - All Sections 5.4 preparation check
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Read Theorem 5.12 in the textbook Links to an external site. .ย Suppose ๐ ๐ โฅ 0 ย for all ๐ โฅ 1 and ๐ ๐ โฅ 0 ย for all ๐ โฅ 1 . a) If โ ๐ = 1 โ ๐ ๐ ย converges and lim ๐ โ โ ๐ ๐ ๐ ๐ = 5 ย then ย โ ๐ = 1 โ ๐ ๐ ย ย [ Select ] might converge or diverge (there is not enough information) converges diverges ย b) If โ ๐ = 1 โ ๐ ๐ ย converges and lim ๐ โ โ ๐ ๐ ๐ ๐ = 0 ย ย then ย โ ๐ = 1 โ ๐ ๐ ย ย [ Select ] diverges converges might converge or diverge (there is not enough information) ย c) If โ ๐ = 1 โ ๐ ๐ ย diverges and lim ๐ โ โ ๐ ๐ ๐ ๐ = 0 ย ย then ย โ ๐ = 1 โ ๐ ๐ ย ย [ Select ] might converge or diverge (there is not enough information) diverges converges ย d) If โ ๐ = 1 โ ๐ ๐ ย diverges and lim ๐ โ โ ๐ ๐ ๐ ๐ = 200 ย ย then ย โ ๐ = 1 โ ๐ ๐ ย ย diverges
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We are given nonnegative sequences a_n and b_n with a_n โฅ 0 and b_n โฅ 0 for all n โฅ 1, and relationships between the series โ b_n and the limit of a_n b_n. We will examine each option carefully.
a) If โ_{n=1}^โ b_n converges and lim_{nโโ} a_n b_n = 5, then โ_{n=1}^โ a_n [Select].
- Since โ b_n converges, we know b_n โ 0 as n โ โ. The condition lim a_n b_n = 5 implies that for large n, a_n behaves roughly like 5 / b_n, which forces a_n to grow without bound because b_n โ 0. In particular, a_n does not approach 0; instead......Login to view full explanation็ปๅฝๅณๅฏๆฅ็ๅฎๆด็ญๆก
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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 ยฑโยฑโ\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 โโ \infty NIF | | if the series equals to โโโโ -\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 aย geometric series โโ๐=0๐๐โn=0โqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐pp-series, where ๐>1p>1p > 1 LP | | Comparison with ๐pp-series, where ๐<1p<1p < 1 [/table] [table] โ๐=1โ(๐+1)7(8โ ๐!)2(2๐)!โn=1โ(n+1)7(8โ n!)2(2n)! \sum_{n=1}^\infty (n+1)^{{7}} \,\, \frac{({8}\cdot n!)^2 }{ (2n)! } | | because โ๐=1โ(1โ6๐)2๐2โn=1โ(1โ6n)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 ยฑโยฑโ\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 โโ \infty NIF | | if the series equals to โโโโ -\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 aย geometric series โโ๐=0๐๐โn=0โqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐pp-series, where ๐>1p>1p > 1 LP | | Comparison with ๐pp-series, where ๐<1p<1p < 1 [/table] [table] โ๐=1โsin(๐9๐)โn=1โsinโก(ฯ9n) \sum_{n=1}^\infty \sin\left( \frac{\pi }{ {9} n } \right) | | because โ๐=1โcos(๐10๐)โn=1โcosโก(ฯ10n) \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 ยฑโยฑโ\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 โโ \infty NIF | | if the series equals to โโโโ -\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 aย geometric series โโ๐=0๐๐โn=0โqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐pp-series, where ๐>1p>1p > 1 LP | | Comparison with ๐pp-series, where ๐<1p<1p < 1 [/table] [table] โ๐=1โ3๐3+15๐12+1โพโพโพโพโพโพโพโ3+๐โn=1โ3n3+15n12+13+n \sum_{n=1}^\infty \frac{{3}n^{3}+1}{{5}\sqrt[{3}]{ n^{12}+1}+n} | | because โ๐=1โ(๐ln๐)24๐โn=1โ(nlnโกn)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 โ{0,1,tan(1/6)}โ{0,1,tanโก(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)tanโก(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 aย geometric series โโ๐=0๐๐โn=0โqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐pp-series, where ๐>1p>1p > 1 LP | | Comparison with ๐pp-series, where ๐<1p<1p < 1 [/table] [table] โ๐=1โ(โ1)๐tan(16๐)โn=1โ(โ1)ntanโก(16n) \sum\limits_{n=1}^{\infty} (-1)^n \tan \left(\frac{1}{{6} n}\right) | | because โ๐=1โ(โ1)๐(1โ7๐)๐โn=1โ(โ1)n(1โ7n)n\sum\limits_{n=1}^{\infty} (-1)^n \left(1-\frac{{7}}{n}\right)^n | | because [/table]
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