Find all values of α such that series ∑ n = 1 ∞ ( 1 n ⋅ sin ⁡ ( 1 / n ) − cos ⁡ ( 1 n ) ) α converges.I used Maclaurin for sin and cos and got: a n = ( 1 1 − 1 3 ! n 2 + … − 1 + 1 2 ! n 2 − 1 4 ! n 4 + … ) α Put it together in one fraction seems to be a hard thing to do.

Question

Find all values of   α such that series      ∑          n      =      1        ∞              (              1                  n          ⋅          sin          ⁡          (          1                      /                    n          )                    −      cos      ⁡              (                  1          n                )            )        α  converges.I used Maclaurin for sin and cos and got:      a    n    =            (              1                  1          −                                    1                              3                !                                  n                  2                                                              +          …                    −      1      +              1                  2          !                      n            2                              −              1                  4          !                      n            4                              +      …      )        α  Put it together in one fraction seems to be a hard thing to do.

Franco Cohen · Accepted Answer

HINT:You are on the right track. Note that      1          1      −              1                  6                      n            2                              +      O              (                  1                      n            4                          )              =  1  +      1          6              n        2              +  O      (          1              n        4              )  Hence, we see that      1          n      sin      ⁡      (      1              /            n      )        −  cos  ⁡  (  1      /    n  )  =      2          3              n        2              +  O      (          1              n        4              )

skynugurq7 · Accepted Answer

Hint  n  sin  ⁡  (      1    n    )  =  1  −      1          6              n        2              (  1  +      ϵ    1    (  n  )  )      1          n      sin      ⁡      (              1        n            )        =  1  +      1          6              n        2              (  1  +      ϵ    2    (  n  )  )  cos  ⁡  (      1    n    )  =  1  −      1          2              n        2              (  1  +      ϵ    3    (  n  )  )thus, when   n  →  +  ∞, the general term of your series       u    n  , satisfies      u    n    ∼  (      2          3              n        2                  )    α  and by the limit comparison test,  ∑      u    n   converges     ⟺      α  &amp;gt;      1    2

Find all values of α such that series <munderover> ∑

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