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MAT137Y1 LEC 20249: Calculus with Proofs (all lecture sections) Pre-Class Quiz 61(13.13 and 13.14)
ๅค้กน้ๆฉ้ข
Let ๐ ย be a continuous function with domain ๐ . Assume ๐ is POSITIVE and DECREASING. Assume lim ๐ฅ โ โ ๐ ( ๐ฅ ) = 0 . Which of the following statements MUST be true? Select all the correct answers.
้้กน
A.โ
๐
=
1
โ
(
โ
1
)
๐
๐
(
๐
)
ย is divergent.
B.IF
โซ
100
โ
๐
(
๐ฅ
)
๐
๐ฅ
ย is convergent, THEN
โ
๐
=
1
โ
๐
(
๐
)
ย is convergent.
C.โ
๐
=
1
โ
(
โ
1
)
๐
๐
(
๐
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ย is convergent.
D.IF
โ
๐
=
1
โ
๐
(
๐
)
is convergent, THEN
โซ
1
โ
๐
(
๐ฅ
)
๐
๐ฅ
ย is convergent.
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We are given a continuous, positive, decreasing function f on the real line with limit zero as x โ โ. We will analyze each statement in light of standard convergence tests.
Option 1: โ_{n=1}^โ (-1)^n f(n) is convergent.
- Because f is positive, decreasing, and tends to 0, the alternating series test (Leibniz criterion) applies: the terms f(n) decrease to 0, so the alternating sum โ (-1)^......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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