Seleft the FIRST correct reason why the given series converges or chose E for diverges.
Aleast one of the answers above is NOT correct. (1 pt) Select the FIRST correct reason why the given series converges. AL A. Convergent geometric series B. Convergent p series C. Integral test D. Comparison with a convergent p series E. Converges by limit comparison test F. Converges by alternating series test 1. LG (cos(17) 2. X 1 In(70) 3. 722 | In(n) M1 72 g 7246 5. ( 1)" 116 (n. 11) (82 1)" | 821 Note: You are...
(1 point) Select the FIRST correct reason why the given series diverges. A. Diverges because the terms don't have limit zero B. Divergent geometric series C. Divergent p series D. Integral test E. Comparison with a divergent p series F Diverges by limit comparison test G. Diverges by alternating series test 1. 2. n ln(n) cos(nT) In(4) 02n 3 (n182 +1)" 2n (1 point) Select the FIRST correct reason why the given series diverges. A. Diverges because the terms don't...
Ch9 Review: Problem 17 Prev Up Next (1 pt) Select the FIRST correct reason why the given series diverges. A. Diverges because the terms don't have limit zero B. Divergent geometric series C. Divergent p series D. Integral test E. Comparison with a divergent p series F Diverges by limit comparison test G. Cannot apply any test done so far in class In(n) 72 2. 72 3. COS 7) Cos nTT In(7) 4. 72 72 12 Note: You can earn...
Series Practice: Problem 4 Previous Problem List Next (3 points) NOTE: Only 3 attempts are allowed for the whole problem Select the FIRST correct reason why the given series diverges. A. Diverges because the terms don't have limit zero B. Divergent geometric series C. Divergent p series D. Integral test E. Comparison with a divergent p series F. Diverges by limit comparison test G. Diverges by alternating series test 1.Zn) 2. o0_ ln(n) cos(nT) nIn(6) 4 Series Practice: Problem 4...
(1 point) Select the FIRST correct reason why the given series converges. A. Convergent geometric series B. Convergent p series C. Comparison (or Limit Comparison) with a geometric or p series D. Alternating Series Test E. None of the above 1. n² + √n n4 – 4 sin?(2n) n2 E 4 (n + 1)(9)" n=1 2n + 2 cos(NT) 16. In(3n)
(3 points) NOTE: Only 3 attempts are allowed for the whole problem Select the FIRST correct reason why the given series diverges A. Diverges because the terms don't have limit zero B. Divergent geometric series C. Divergent p series D. Integral test E. Comparison with a divergent p series F. Diverges by limit comparison test G. Diverges by alternating series test cos(nT) In(5) 2 1t 00 n(n) 4 1t 1t n In(n) (3 points) NOTE: Only 3 attempts are allowed...
(1 point) Select the FIRST correct reason why the given series converges. A. Convergent geometric series B. Convergent p series C. Comparison with a geometric or p series D. Alternating Series Test E. None of the above 1. Cos(17) (ln(6n) (n + 1)(80)" (n + 2)92n n² | 6. § (-1)",
Select the FIRST correct reason why the given series diverges. A. Divergent p-series B. Divergent geometric series C. Comparison with a divergent p-series D. Diverges because the terms don't have limit zero E. Integral test D 1. In(n) N=3 In(n) A 2. n UM IM UMUM8 A !!! 3. E 4. 1 n ln(n) n=3
Not secure csuwebworks.website/webwork2/MATH-221-06-Spring-20/P5-221/6/ YouTube Maps ECOLO G UE 2 of the questions remain unanswered. CHICO (1 point) Decide whether each of the following series converges. If a given series converges, compute its sum. Otherwise, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, and DIV otherwise. (e13 – c126-) (sin(6n) - sin(6(n + 1))) Note: In order to get credit for this problem all answers must be correct Type here to search At least one...
To test the series e 2n for convergence, you can use the Integral Test. (This is also a geometric series, so we could n=1 also investigate convergence using other methods.) Find the value of e-24 dx = Preview Ji What does this value tell you about the convergence of the series e-2n? the series definitely diverges the series might converge or diverge: we need more information the series definitely converges Compute the value of the following improper integral, if it...