Comparison tests Use the Comparison Test or Limit Comparison Test to determine whether the following series converge.
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- Find the interval of convergence for the given power series. (x - 4)" Σ n(- 9)" n=1 The series is convergent from x = left end included (enter Y or N): to x = right end included (enter Y or N): M C ㅈ # $ A de L % 5 6 D 8 7 8 9 #arrow_forwardselect the correct answer and explain step by steparrow_forwardState whether it converges or diverges. Justify it using either a basic divergence, integral, basic comparison, limit comparison, alternating series, root or ratio testarrow_forward
- OSelect 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 (n + 1)(8)" 1. 32n (-1)" 2. 7n + 5 n=1 00 3. (-1)"- n+3 7(7)" A 4. 2n sin (4n) 5. n? (-1)" In(e") n² cos(na) 00 6.arrow_forward4n Does the series 2 converge or diverge? 8n n=1 1+ earrow_forward00 Does the seriesE(- 1n+12+n° n4 converge absolutely, converge conditionally, or diverge? n= 1 Choose the correct answer below and, if necessary, fill in the answer box to complete your choice. O A. The series converges absolutely per the Comparison Test with > 00 n4 n= 1 B. The series diverges because the limit used in the Ratio Test is not less than or equal to 1. OC. The series converges conditionally per the Alternating Series Test and the Comparison Test with n= 1 D. The series converges absolutely because the limit used in the nth-Term Test is E. The series diverges because the limit used in the nth-Term Test does not exist. O F. The series converges conditionally per the Alternating Series Test and because the limit used in the Ratio Test isarrow_forward
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