Convergence and limit of ∑ j = 1 n sin ⁡ ( j − 1 n ) { cos ⁡ ( j n ) − cos ⁡ ( j − 1 n ) }

Question

Convergence and limit of       ∑          j      =      1        n    sin  ⁡      (                  j        −        1            n        )        {    cos    ⁡          (              j        n            )        −    cos    ⁡          (                        j          −          1                n            )        }

mar1nerne · Accepted Answer

We have:      lim          n      →      +      ∞            ∑          j      =      1              n        sin  ⁡      (                  j        −        1            n        )        [    cos    ⁡          (              j        n            )        −    cos    ⁡          (                        j          −          1                n            )        ]    =                    sin        ⁡        (        2        )        −        2            4        .Sketch of proof: we have:  cos  ⁡      (          j      n        )    −  cos  ⁡      (                  j        −        1            n        )    =      ∫                            j          −          1                n                            j        n              (  −  sin  ⁡  x  )    d  x  =  −      1    n      sin  ⁡      (                  j        −        1            n        )    +  O      (          1              n        2              )    ,hence our limit is the same as:  −      lim          n      →      +      ∞            ∑          j      =      1              n            1    n        sin    2    ⁡      (                  j        −        1            n        )    =  −      ∫          0              1            sin    2    ⁡  (  x  )    d  x  .

Convergence and limit of <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"

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