题目
MAT136H5 S 2025 - All Sections 5.6 preparation check
多重下拉选择题
Let's use the Ratio Test to find the convergence of the series ∞ ∑ n=1 3n n2 . a) What is an in this example? [ Select ] 2 3^n n^2 3^n / n^2 3 b) Evaluate the limit lim n→∞| an+1 an |. [ Select ] 2/3 0 1 3/2 3 infinity c) Your answer for (b) means that the series ∞ ∑ n=1 3n n2 : [ Select ] The ratio test does not provide any information Diverges Converges absolutely Converges conditionally
查看解析
标准答案
Please login to view
思路分析
The problem asks us to analyze the convergence of the series with general term a_n = 3^n / n^2 using the Ratio Test, and it provides three sub-questions (a, b, c).
First, for part (a) we need to identify the correct expression for a_n in this example from the given options. The series is the sum over n of 3^n / n^2, so the correct a_n ......Login to view full explanation登录即可查看完整答案
我们收录了全球超50000道考试原题与详细解析,现在登录,立即获得答案。
类似问题
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∞(nlnn)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]
更多留学生实用工具
希望你的学习变得更简单
加入我们，立即解锁 海量真题 与 独家解析，让复习快人一步！