Determine convergence / divergence of ∑ sin ⁡ π n 2 .Let a n = sin ⁡ π n 2 .I attempted the integral test but on the interval [ 1 , 2 ) it is increasing and decreasing on ( 2 , ∞ ). So the integral test in only applicable for the decreasing part. + the integral computation seems to lead to 3 pages of steps...I believe the comparison test would be the most reasonable test, graphically I observed that a n behaves like b n = 1 / n when n is large. ∑ b n = ∞; but, since b n &gt; a n ⟹ inconclusive for divergence/convergence.I attempted the limit comparison, and ratio test, but inconclusive. I am uncertain if I am doing them properly.How could I bound below a n to proceed with the comparison test? Is there a more appropriate method? How would you proceed?

Question

Determine convergence / divergence of   ∑  sin  ⁡      π          n      2      .Let       a    n    =  sin  ⁡      π          n      2      .I attempted the integral test but on the interval   [  1  ,      2    ) it is increasing and decreasing on   (      2    ,  ∞  ). So the integral test in only applicable for the decreasing part. + the integral computation seems to lead to 3 pages of steps...I believe the comparison test would be the most reasonable test, graphically I observed that       a    n   behaves like       b    n    =  1      /    n when n is large.  ∑      b    n    =  ∞; but, since       b    n    &amp;gt;      a    n      ⟹   inconclusive for divergence/convergence.I attempted the limit comparison, and ratio test, but inconclusive. I am uncertain if I am doing them properly.How could I bound below       a    n   to proceed with the comparison test? Is there a more appropriate method? How would you proceed?

Alexia Mata · Accepted Answer

Step 1It is not true that       a    n   behaves like 1/n for large n, because      |    sin  ⁡  (  x  )      |    ≤      |    x      |  for all x (and in fact,       |    sin  ⁡  x      |    ≈      |    x      |   for values of x near zero). In particular, this means that      a    n    ≤  π      /        n    2    ..Step 2If you prefer, the limit comparison test can be applied with       b    n    =  π      /        n    2   and using the sharpness of the small angle approximation mentioned above.

Trevor Rush · Accepted Answer

Explanation:Converges because   sin  ⁡      (          π              n        2              )    ≤      π          n      2       and       ∑          n      =      1              ∞            1          n      2        =            π      2        6  .

Determine convergence / divergence of sum sin pi/n^2

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