Absolute convergence of the series ∑ n = 1 ∞ ( − 1 ) n ln ⁡ ( cos ⁡ ( 1 n ) )

Question

Absolute convergence of the series       ∑          n      =      1              ∞        (  −  1      )    n    ln  ⁡      (    cos    ⁡          (              1        n            )        )

zastenjkcy · Accepted Answer

Just use an asymptotic expansion:  ln  ⁡  (  cos  ⁡  (  1      /    n  )  )  =  ln  ⁡  (  1  −  1      /    2      n    2    +  o  (  1      /        n    2    )  )  ∼  −      1          2              n        2            By the way, for   0  &amp;lt;  x  &amp;lt;      π    2  , we have      d          d      x        ln  ⁡  (  cos  ⁡  (  x  )  )  =  −            sin      ⁡      x              cos      ⁡      x        &amp;lt;  0so that   −  ln  ⁡  (  cos  ⁡  (  1      /    n  )  ) is decreasingly converging to 0. Hence, the alternating series test does apply.

Celeste Barajas · Accepted Answer

As the other answerer mentioned,   ln  ⁡  (  cos  ⁡  (  x  )  )  ∼  −                    x        2            2      . This can stated formally and proved using L&#039;hopital&#039;s rule:      lim          x      →      0                          ln        ⁡        (        cos        ⁡        (        x        )        )                    x        2              =      lim          x      →      0                          (        ln        ⁡        (        cos        ⁡        (        x        )        )                  )          ′                            (                  x          2                          )          ′                      =      lim          x      →      0        −                    sin        ⁡        (        x        )                    2        x        cos        ⁡        (        x        )              =      lim          x      →      0                          sin        ⁡        (        x        )            x        ×      lim          x      →      0        −            1              2        cos        ⁡        (        x        )              =  −            1      2      As a result,      lim          n      →      ∞                                    |                log        ⁡        cos        ⁡        (                              1            n                          )                  |                                      1                      n            2                                =            1      2      So the series converges absolutely by the limit comparison test, since we know       ∑          n      =      1              ∞                  1              n        2             converges.

Absolute convergence of the series sum_(n=1)^infty (-1)^n ln(cos(1/n))

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