Find the value of : lim n → ∞ ∏ k = 1 n cos ⁡ ( k a n n )

Question

Find the value of :       lim          n      →      ∞            ∏          k      =      1        n    cos  ⁡      (                  k        a                    n                  n                      )

Alison Mcgrath · Accepted Answer

Let  f  (  n  )  =      ∏          k      =      1        n    cos  ⁡      (                            k          a                          n                      n                                )    g  (  n  )  =  log  ⁡  (  f  (  n  )  )  =      ∑          k      =      1              n        log  ⁡      (    cos    ⁡          (                                    k            a                                n                          n                                          )        )    =      ∑          k      =      1              n        log  ⁡      (    1    −                                        (                                                            k                  a                                                  n                                      n                                                                        )                    2                2              +          O              (                                    k            4                                n            6                              )        )    log  ⁡      (    1    −                                        (                                                            k                  a                                                  n                                      n                                                                        )                    2                2              +          O              (                                    k            4                                n            6                              )        )    =  −      (                                        (                                                            k                  a                                                  n                                      n                                                                        )                    2                2              +          O              (                                    k            4                                n            6                              )        )    +      O        (                            k          4                          n          6                      )        ∑          k      =      1              n        log  ⁡      (    1    −                                        (                                                            k                  a                                                  n                                      n                                                                        )                    2                2              +          O              (                                    k            4                                n            6                              )        )    =      ∑          k      =      1              n            (    −                                        (                                                            k                  a                                                  n                                      n                                                                        )                    2                2              +          O              (                                    k            4                                n            6                              )        )      =  −                    a        2                    2                  n          3                                        n        (        n        +        1        )        (        2        n        +        1        )            6        +      O    (  1      /    n  )      lim          n      →      ∞            ∑          k      =      1              n        log  ⁡      (    1    −                                        (                                                            k                  a                                                  n                                      n                                                                        )                    2                2              +          O              (                                    k            4                                n            6                              )        )    =  −                    a        2            6      Hence,      ∏          k      =      1              ∞        cos  ⁡      (                            k          a                          n                      n                                )    =  exp  ⁡  (  −      a    2        /    6  )The solution you have   1  −      a    2        /    6 is a first order approximation to   exp  ⁡  (  −      a    2        /    6  ) since  exp  ⁡  (  x  )  =  1  +  x  +      O    (      x    2    )

Find the value of : lim_(n => infty) prod_(k=1)^n cos((ka)/(n sqrt(n)))

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