Select the FIRST correct reason why the given deries converges. A. Convergent geometric series B. Convergent p series C. Comparison (of Limit Comparison) with a geometric or p series D. Converges by alternating series test 1. ∑n=1∞sin2(3n)n2 2. ∑n=1∞(−1)nln⁡(en)n4cos⁡(nπ) 3. ∑n=1∞cos⁡(nπ)ln⁡(4n) 4. ∑n=1∞2(4)n62n 5. ∑n=1∞(−1)n2n+6 6. ∑n=1∞(−1)nnn+4

Question

Select the FIRST correct reason why the given deries converges.  A. Convergent geometric series  B. Convergent p series  C. Comparison (of Limit Comparison) with a geometric or p series  D. Converges by alternating series test  1. ∑n=1∞sin2(3n)n2  2. ∑n=1∞(−1)nln⁡(en)n4cos⁡(nπ)  3. ∑n=1∞cos⁡(nπ)ln⁡(4n)  4. ∑n=1∞2(4)n62n  5. ∑n=1∞(−1)n2n+6  6. ∑n=1∞(−1)nnn+4

Bob Huerta · Accepted Answer

1) an=sin2(3n)n2, an=1n2  Converges by comparsion with p-series  → option (C)  2) an=(−1)nln⁡(en)n4cos⁡(nπ)=(−1)nnln⁡(e)n4(−1)n=1n3  → convergent p-series  → option (B)

RizerMix · Accepted Answer

5) an=(−1)n2n+6 → convergent by alternating series test → option (D) 6) an=(−1)nnn+4 → diverges by alternating series test → option (D)

Jeffrey Jordon · Accepted Answer

3) an=cos⁡(nπ)ln⁡(4n)=(−1)nln⁡(4n) → Converges by alternative series test → option (D) 4) an=24n62n=2(436)n=2(19)n → It is convergent Geometric series → option (A)

Select the FIRST correct reason why the given deries converges. A.

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