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kramberol 2022-07-10 Answered
Convergence / divergence examination k = 1 sin k ( a ) k
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Answers (1)

talhekh
Answered 2022-07-11 Author has 15 answers
If sin ( a ) = 0, then the series obviously converges.
If sin ( a ) = 1, then the series rewrites 1 k which diverges.
If sin ( a ) = 1, then the series rewrites ( 1 ) k k which converges by Alternating Test.
If | sin ( a ) | < 1 and sin ( a ) 0, then one has
| sin k + 1 ( a ) / ( k + 1 ) sin k ( a ) / k | = | sin ( a ) | × k + 1 k | sin ( a ) |
Because | sin ( a ) | < 1, then by the ratio test, the series converges.
In conclusion, the series converges iff sin ( a ) 1, i.e.
the series converges iff  a π 2  mod  2 π
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