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MAT137Y1 LEC 20249: Calculus with Proofs (all lecture sections) Pre-Class Quiz 59 (13.8 and 13.9)

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Letย  ๐‘† = โˆ‘ ๐‘› = 0 โˆž ๐‘Ž ๐‘› ย  be a series.ย  Which of the following statements must be true? Select all the correct answers.

้€‰้กน
A.IF ย  lim ๐‘› โ†’ โˆž ๐‘Ž ๐‘› โ‰  0 ย ย ,ย  THENย  the series S is DIVERGENT.
B.IF the series S is DIVERGENT,ย  THEN ย  lim ๐‘› โ†’ โˆž ๐‘Ž ๐‘› = 0 .
C.IF ย  lim ๐‘› โ†’ โˆž ๐‘Ž ๐‘› = 0 ,ย  THENย  the series S is CONVERGENT.
D.IFย  โˆ‘ ๐‘› = 1 โˆž ๐‘Ž ๐‘› is DIVERGENT, THEN โˆ‘ ๐‘› = 1 โˆž ๐‘Ž ๐‘› = โˆž
E.IFย  โˆ‘ ๐‘› = 1 โˆž ๐‘Ž ๐‘› is DIVERGENT, THEN lim ๐‘› โ†’ โˆž ๐‘Ž ๐‘› โ‰  0 ย .
F.IF the series S is CONVERGENT,ย  THEN ย  lim ๐‘› โ†’ โˆž ๐‘Ž ๐‘› = 0 .
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We start by restating the problem: we have a series S = sum from n = 0 to โˆž of a_n, and we must determine which of the given statements must be true. Option 1: 'IF lim_{nโ†’โˆž} a_n โ‰  0, THEN the series S is DIVERGENT.' - This is a standard necessary condition for convergence: if a_n does not tend to 0, the series cannot converge. Therefore, this statement must be true. Option 2: 'IF the series S is DIVERGENT, THE......Login to view full explanation

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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 ยฑโˆžยฑโˆž\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 โˆžโˆž \infty NIF | | if the series equals to โˆ’โˆžโˆ’โˆž -\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 โˆ‘โˆž๐‘›=0๐‘ž๐‘›โˆ‘n=0โˆžqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐‘pp-series, where ๐‘>1p>1p > 1 LP | | Comparison with ๐‘pp-series, where ๐‘<1p<1p < 1 [/table] [table] โˆ‘๐‘›=1โˆž(๐‘›+1)7(8โ‹…๐‘›!)2(2๐‘›)!โˆ‘n=1โˆž(n+1)7(8โ‹…n!)2(2n)! \sum_{n=1}^\infty (n+1)^{{7}} \,\, \frac{({8}\cdot n!)^2 }{ (2n)! } | | because โˆ‘๐‘›=1โˆž(1โˆ’6๐‘›)2๐‘›2โˆ‘n=1โˆž(1โˆ’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 ยฑโˆžยฑโˆž\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 โˆžโˆž \infty NIF | | if the series equals to โˆ’โˆžโˆ’โˆž -\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 โˆ‘โˆž๐‘›=0๐‘ž๐‘›โˆ‘n=0โˆžqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐‘pp-series, where ๐‘>1p>1p > 1 LP | | Comparison with ๐‘pp-series, where ๐‘<1p<1p < 1 [/table] [table] โˆ‘๐‘›=1โˆžsin(๐œ‹9๐‘›)โˆ‘n=1โˆžsinโก(ฯ€9n) \sum_{n=1}^\infty \sin\left( \frac{\pi }{ {9} n } \right) | | because โˆ‘๐‘›=1โˆžcos(๐œ‹10๐‘›)โˆ‘n=1โˆžcosโก(ฯ€10n) \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 ยฑโˆžยฑโˆž\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 โˆžโˆž \infty NIF | | if the series equals to โˆ’โˆžโˆ’โˆž -\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 โˆ‘โˆž๐‘›=0๐‘ž๐‘›โˆ‘n=0โˆžqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐‘pp-series, where ๐‘>1p>1p > 1 LP | | Comparison with ๐‘pp-series, where ๐‘<1p<1p < 1 [/table] [table] โˆ‘๐‘›=1โˆž3๐‘›3+15๐‘›12+1โ€พโ€พโ€พโ€พโ€พโ€พโ€พโˆš3+๐‘›โˆ‘n=1โˆž3n3+15n12+13+n \sum_{n=1}^\infty \frac{{3}n^{3}+1}{{5}\sqrt[{3}]{ n^{12}+1}+n} | | because โˆ‘๐‘›=1โˆž(๐‘›ln๐‘›)24๐‘›โˆ‘n=1โˆž(nlnโกn)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 โˆ‰{0,1,tan(1/6)}โˆ‰{0,1,tanโก(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)tanโก(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 โˆ‘โˆž๐‘›=0๐‘ž๐‘›โˆ‘n=0โˆžqn \sum_{n=0}^{\infty}q^n CH | | Comparison with the harmonic series AH | | Comparison with the alternating harmonic series [/table] [table] P | | Comparison with ๐‘pp-series, where ๐‘>1p>1p > 1 LP | | Comparison with ๐‘pp-series, where ๐‘<1p<1p < 1 [/table] [table] โˆ‘๐‘›=1โˆž(โˆ’1)๐‘›tan(16๐‘›)โˆ‘n=1โˆž(โˆ’1)ntanโก(16n) \sum\limits_{n=1}^{\infty} (-1)^n \tan \left(\frac{1}{{6} n}\right) | | because โˆ‘๐‘›=1โˆž(โˆ’1)๐‘›(1โˆ’7๐‘›)๐‘›โˆ‘n=1โˆž(โˆ’1)n(1โˆ’7n)n\sum\limits_{n=1}^{\infty} (-1)^n \left(1-\frac{{7}}{n}\right)^n | | because [/table]

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