Use the comparison or limit comparison test to decide if the following series converge. ∑n=1∞4−sin⁡n2n+1 For each series which converges, give an approximation of its sum, together with an error estimate, as follows. First calculate the sum s5 of the first 5 terms, then estimate the "tail" ∑n=6∞an by comparing it with an appropriate improper integral or geometric series.

Question

Use the comparison or limit comparison test to decide if the following series converge.   ∑n=1∞4−sin⁡n2n+1  For each series which converges, give an approximation of its sum, together with an error estimate, as follows. First calculate the sum s5 of the first 5 terms, then estimate the &quot;tail&quot; ∑n=6∞an by comparing it with an appropriate improper integral or geometric series.

SlabydouluS62 · Accepted Answer

∑n=1∞4−sin⁡n2n+1  Since 4−sin⁡n2n+1&amp;lt;32n (since the maximum value of sin⁡n is 1)  So ∑n=1∞4−sin⁡n2n+1&amp;lt;∑n=1∞32n  Since ∑n=1∞32n is convergent so ∑n=1∞4−sin⁡n2n+1 is convergent  So the approximate value  ∑n=1∞32n=3(12+122+123…)+  =3{(1+12+122+123…)−1}  =3{(1−12)−1−1}  =3{(12)−1−1}  =3(2−1)  =3

chumants6g · Accepted Answer

Use the root test to determine the convergence of the series:  ∑n∞4−sin⁡(n)2n+1  Recall the statement of the root test.  The convergence of a series ∑k∞ak can be determined by examining the quantity ρ=limk→∞|ak|1k using the following comparisons:  If ρ&amp;lt;1, the series converges absolutely.  If ρ&amp;gt;1 or ρ=∞, the series diverges.  If ρ=1, the series may converge or diverge and the test is inconclusive.  Check the convergence of the series by taking the limit of |4−sin⁡(n)2n+1|1n.  Take the limit of the n-th root of the absolute value of the n-th term.  limn→∞|4−sin⁡(n)2n+1|1n=12  Compare the result to 1.  Since the limit is less than 1, the series ∑n∞4−sin⁡(n)2n+1 converges absolutely:  Answer: The series is absolutely convergent by the root test.

user_27qwe · Accepted Answer

The first answer is clearer

Use the comparison or limit comparison test to decide if the

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