Using either the Direct or Limit Comparison Tests, determine if ∑ n = 1 ∞ 1 + sin 2 ⁡ ( n ) 3 n is convergent or divergent.

Question

Using either the Direct or Limit Comparison Tests, determine if       ∑          n      =      1              ∞                  1      +              sin                  2                    ⁡      (      n      )              3      n       is convergent or divergent.

erlentzed · Accepted Answer

Since   1  ≤  1  +      sin    2    ⁡  (  n  )  ≤  2, we can compare to the geometric seriesIn fact, this is actually the sum of three geometric series:                              ∑                      n            =            1                    ∞                                      1            +                          sin              2                        ⁡            (            n            )                                3            n                                              =                  ∑                      n            =            1                    ∞                                      1            −                          1              4                        (                          e                              2                i                n                                      −            2            +                          e                              −                2                i                n                                      )                                3            n                                                            =                  3          2                          ∑                      n            =            1                    ∞                          1                      3            n                          −                  1          2                          R          e                          (                      ∑                          n              =              1                        ∞                                              e                              2                i                n                                                    3              n                                )                                                  =                  3          2                ⋅                              1                          /                        3                                1            −            1                          /                        3                          −                  1          2                ⋅                  R          e                          (                                                    e                                  2                  i                                                            /                            3                                      1              −                              e                                  2                  i                                                            /                            3                                )                                                  =                  3          4                −                  1          2                ⋅                  R          e                          (                                                    e                                  2                  i                                                            /                            3                                      1              −                              e                                  2                  i                                                            /                            3                                ⋅                                    1              −                              e                                  −                  2                  i                                                            /                            3                                      1              −                              e                                  −                  2                  i                                                            /                            3                                )                                                  =                  3          4                −                  1          2                ⋅                  R          e                          (                                                    e                                  2                  i                                                            /                            3              −              1                              /                            9                                      10                              /                            9              −              2              cos              ⁡              (              2              )                              /                            3                                )                                                  =                  3          4                −                  1          4                                      3            cos            ⁡            (            2            )            −            1                                5            −            3            cos            ⁡            (            2            )                                                            =                              4            −            3            cos            ⁡            (            2            )                                5            −            3            cos            ⁡            (            2            )                                                            ≐        0.83996006708282            As estimated, the sum is between 1/2 and 1

Using either the Direct or Limit Comparison Tests, determine if sum_(n=1)^infty (1+sin^2(n))/3^n

Answered question

Answer & Explanation

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