Solution 03 23 2016 at 03 00am EDT 1 pt Consider the series n 1 n 7n2 2 4 Determine whether the series converges and if it converges determine its
Solution at am EDT pt Consider the series n n n Determine whether the series converges and
Solution at am EDT pt Consider the series n n n Determine whether
pt Consider the series n n n Determine whether the series converges and if it converges determine its
Solution at am EDT pt Consider the series n n n
Determine whether the series converges and if it converges determine its
Solution at am EDT pt Consider the series
Solution at am EDT
(Solution) 03/23/2016 at 03:00am EDT. (1 pt) Consider the series n=1 n 7n2 + 2 4. Determine whether the series converges, and if it converges, determine its...

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Hi, i need help please. answer questions 2, 5 and 10 only, thanks, show work.Mohammed Yaaseen Gomdah E WeBWorK assignment due : 03/23/2016 at 03:00am EDT. 1. (1 pt) Consider the series ? ? n = 1 n ? 7 n 2 + 2 Determine whether the series converges, and if it converges, determine its value. If the sum does not converge, enter DNE . Value (or DNE ): 2. (1 pt) Compute the value of the following improper inte- gral. If it converges, enter its value. Enter in?nity if it diverges to ? , and -in?nity if it diverges to - ? . Otherwise, enter di- verges. Z ? 1 2ln ( x ) x 8 dx = Does the series ? ? n = 1 2ln ( n ) n 8 converge or diverge? ? 3. (1 pt) Determine whether the series is convergent or diver- gent. If convergent, ?nd the sum; if divergent, enter div . ? ? n = 1 n ? n 2 + 17 Answer: 4. (1 pt) Evaluate Z ? 1 7 x 2 e - x 3 Answer: Determine whether the following series congerges. ? ? n = 1 7 n 2 e - n 3 Enter C if series is convergent, D if series is divergent: 5. (1 pt) Test each of the following series for convergence by the Integral Test. If the Integral Test can be applied to the series, enter CONV if it converges or DIV if it diverges. If the integral test cannot be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the Integral Test cannot be applied to it, then you must enter NA rather than CONV.) 1. ? ? n = 1 n + 1 ( - 6 ) n 2. ? ? n = 1 ln ( 4 n ) n 3. ? ? n = 1 7 n ln ( 4 n ) 4. ? ? n = 1 7 n ( ln ( 4 n )) 6 5. ? ? n = 1 ne 3 n 6. (1 pt) Determine the convergence or divergence of the fol- lowing series. ? ? n = 1 ? 2 n + 1 n 2 • A. convergent • B. divergent 7. (1 pt) Determine whether the series is convergent or diver- gent. If convergent, ?nd the sum; if divergent, enter div . ? ? n = 1 n n + 4 Answer: 8. (1 pt) Match each of the following with the correct state- ment. C stands for Convergent, D stands for Divergent. 1. ? ? n = 1 ln ( n ) 7 n 2. ? ? n = 1 4 n 7 - 49 3. ? ? n = 1 4 n ( n + 5 ) 4. ? ? n = 1 4 + 7 n 4 + 9 n 5. ? ? n = 1 1 4 + 4 ? n 4 9. (1 pt) Test each of the following series for convergence by either the Comparison Test or the Limit Comparison Test. If either test can be applied to the series, enter CONV if it con- verges or DIV if it diverges. If neither test can be applied to the series, enter NA. (Note: this means that even if you know a given series converges by some other test, but the comparison tests cannot be applied to it, then you must enter NA rather than CONV.) 1. ? ? n = 1 ( ln ( n )) 5 n + 8 2. ? ? n = 1 cos ( n ) ? n 6 n + 9 1

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