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MAT136H5 S 2025 - All Sections 5.6 preparation check
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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
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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 explanationLog in for full answers
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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 ±∞±∞\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]
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