Evaluating ∑ n = 0 ∞ ( 1 / 2 ) n n ! ( H n − H n − 1 / 2 )

Question

Evaluating       ∑          n      =      0        ∞              (      1              /            2              )        n                    n      !        (      H    n    −      H          n      −      1              /            2        )

Allyson Gonzalez · Accepted Answer

From the integral representation      H    p    =      ∫    0    1              1      −              t        p                    1      −      t          d  twe have                              H          n                −                  H                      n            −            1                          /                        2                                              =                  ∫          0          1                                                    t                              n                −                1                                  /                                2                                      −                          t              n                                            1            −            t                                  d        t                                          =                  ∫          0          1                                                    t                              n                −                1                                  /                                2                                      (            1            −                          t                        )                                1            −            t                                  d        t                                          =        2                  ∫          0          1                                      u                          2              n                                            1            +            u                                  d        u            The proposed series can be written as                    S                            =                  ∑                      n            =            0                    ∞                                      (            1                          /                        2                          )              n                                            n            !                          (                  H          n                −                  H                      n            −            1                          /                        2                          )                                          =        2                  ∑                      n            =            0                    ∞                                      (            1                          /                        2                          )              n                                            n            !                                    ∫          0          1                                      u                          2              n                                            1            +            u                                  d        u            by swapping integration and summation                    S                            =        2                  ∫          0          1                                      d            u                                1            +            u                                    ∑                      n            =            0                    ∞                                      (            1                          /                        2                          )              n                                            n            !                                    u                      2            n                                                            =        2                  ∫          0          1                                      d            u                                (            1            +            u            )                          1              −                              u                2                                                                                      =                  ∫          0          1                          v                      −            1                          /                        2                                  d        v                                          =        2            (the last integral was obtained by changing   u  =  (  1  −  v  )      /    (  1  +  v  ))

Evaluating <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n =

Answered question

Answer & Explanation

New Questions in Calculus 2