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Name: MAC 0. Does the following series meet the requirements of the Alternating Series test? Why...
To which of the following series can we apply the alternating series test ? 0 *...
To which of the following series can we apply the alternating series test ? 0 * cos(in) 21+ m2 0° 1 – 2 sinº(n) n=1 1+n2 n=1 og sin(1/n) 1 + n2 n=1 Х 1 n=1 (1 + n2)(cos(n))
(1 pt) Test each of the following series for convergence by either the Comparison Test or...
(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 converges 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...
(1 point) Select the FIRST correct reason why the given series diverges. A. Diverges because the ...
(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...
(1 point) Select the FIRST correct reason why the given series converges. A. Convergent geometric series...
(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)",
Question 11 0/5 points n+1 satisfies all requirements of the Alternating Series Test. (You don't It...
Question 11 0/5 points n+1 satisfies all requirements of the Alternating Series Test. (You don't It 2n=1 have to check that - trust me on this one.) (2n+1) (a) Use a calculator to evaluate the partial sum S3 of this series. Give the answer rounded to four decimal places. (b) Estimate the error of using S3 as an approximation to the sum of the series, i.e. estimate the remainder R3. Recall that the remainder estimate of the Alternating Series Test...
1. (Alternating Series Test.) This shows that for this particular sort of alternating series, the error...
1. (Alternating Series Test.) This shows that for this particular sort of alternating series, the error in approximating the infinite sum by a partial sum is at most the first omitted term. Suppose that aj > a2 > a3 > ... > 0 and that limnyoo An = 0. Let sn = {k=1(-1)kak. (a) Prove that if n > m > 0 then |sn – Sm! < am+1. (b) Prove that 2-1(-1)kak converges and that, for all n > 0,...
7. Use the Alternating Series Test to determine the convergence or divergence of the series
7. Use the Alternating Series Test to determine the convergence or divergence of the series a) \(\sum_{n=1}^{\infty} \frac{(-1)^{n} \sqrt{n}}{2 n+1}\)b) \(\sum_{n=1}^{\infty} \frac{(-1)^{n} n}{2 n-1}\)8. Use the Ratio Test or the Root Test to determine the convergence or divergence of the seriesa) \(\sum_{n=0}^{\infty}\left(\frac{4 n-1}{5 n+7}\right)^{n}\)b) \(\sum_{n=0}^{\infty} \frac{\pi^{n}}{n !}\)
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 bec...
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 pt) Test each of the following series for convergence by either the Comparison Test or...
(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 converges 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...
The series converges by the Alternating Series Test. Use Theorem 9.9: Error Bounds for Alternatin...
The series converges by the Alternating Series Test. Use Theorem
9.9: Error Bounds for Alternating Series to find how many terms
give a partial sum, Sn, within 0.01 of the sum, S, of
the series.
-1 I n Theorem 9.9: Error Bounds for Alternating Series Let n = Σ Suppose that 0 < an+1 < an for all n and limn-too an-0. Then (- 1)i-lai be the nth partial sum of an alternating series and let S = lim Sn....
To which of the following series can we apply the alternating series test ? 0 * cos(in) 21+ m2 0° 1 – 2 sinº(n) n=1 1+n2 n=1 og sin(1/n) 1 + n2 n=1 Х 1 n=1 (1 + n2)(cos(n))
(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 converges 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...
(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...
(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)",
Question 11 0/5 points n+1 satisfies all requirements of the Alternating Series Test. (You don't It 2n=1 have to check that - trust me on this one.) (2n+1) (a) Use a calculator to evaluate the partial sum S3 of this series. Give the answer rounded to four decimal places. (b) Estimate the error of using S3 as an approximation to the sum of the series, i.e. estimate the remainder R3. Recall that the remainder estimate of the Alternating Series Test...
1. (Alternating Series Test.) This shows that for this particular sort of alternating series, the error in approximating the infinite sum by a partial sum is at most the first omitted term. Suppose that aj > a2 > a3 > ... > 0 and that limnyoo An = 0. Let sn = {k=1(-1)kak. (a) Prove that if n > m > 0 then |sn – Sm! < am+1. (b) Prove that 2-1(-1)kak converges and that, for all n > 0,...
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 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 converges 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...
The series converges by the Alternating Series Test. Use Theorem
9.9: Error Bounds for Alternating Series to find how many terms
give a partial sum, Sn, within 0.01 of the sum, S, of
the series.
-1 I n Theorem 9.9: Error Bounds for Alternating Series Let n = Σ Suppose that 0 < an+1 < an for all n and limn-too an-0. Then (- 1)i-lai be the nth partial sum of an alternating series and let S = lim Sn....
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