Calculate the sum ∑ n = 1 ∞ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ⋅ 2 n where ( 2 n − 1 ) ! ! = 1 ⋅ 3 ⋯ ( 2 n − 1 ), ( 2 n ) ! ! = 2 ⋅ 4 ⋯ 2 n

Question

Calculate the sum            ∑              n        =        1                    ∞                            (        2        n        −        1        )        !        !                    (        2        n        )        !        !        ⋅                  2          n                    where   (  2  n  −  1  )  !  !  =  1  ⋅  3  ⋯  (  2  n  −  1  ),   (  2  n  )  !  !  =  2  ⋅  4  ⋯  2  n

rideonthebussp · Accepted Answer

First note that  (  2  n  −  1  )  !  !  =            (      2      n      )      !                      2        n            n      !      and   (  2  n  )  !  !  =      2    n    n  !            (      2      n      −      1      )      !      !              (      2      n      )      !      !      ⋅              2        n              =            (      2      n      )      !                      2        n            n      !      ⋅              2        n            n      !      ⋅              2        n              =      1          2              3        n                                (                            2        n            n                      )                .Now use the power series                    (1)                              1                      1            −            4            x                          =                  ∑                      n            ≥            0                                                              (                                                          2              n                        n                                              )                                                x          n                        ,            the generating function for the central binomial coefficients.                     (                            2        n            n                      )              ≤      4    n   for   n  ≥  1, so (1) certainly converges at   x  =      1    8  , and we have      2    =      1          1      −      1              /            2        =      ∑          n      ≥      0                          (                            2        n            n                      )                        (              1        8            )        n    =      ∑          n      ≥      0            1          2              3        n                                (                            2        n            n                      )              =  1  +      ∑          n      ≥      1                  (      2      n      −      1      )      !      !              (      2      n      )      !      !      ⋅              2        n                .

wanaopatays · Accepted Answer

Since   (  2  n  )  !  !  =      2    n    n  ! and   (  2  n  −  1  )  !  !  =  (  2  n  −  1  )  (  2  n  −  3  )  ⋯  5  ⋅  3  ⋅  1                              ∑                      n            =            1                    ∞                                      (            2            n            −            1            )            !            !                                (            2            n            )            !            !                          2              n                                                          =                  ∑                      n            =            1                    ∞                                      (            2            n            −            1            )            (            2            n            −            3            )            ⋯            5            ⋅            3            ⋅            1                                              2              n                        n            !                                    1                      2            n                                                            =                  ∑                      n            =            1                    ∞                                                    (              n              −                              1                2                            )                                      (              n              −                              3                2                            )                        ⋯                          (                              3                2                            )                                      (                              1                2                            )                                            n            !                                    1                      2            n                                                            =                  ∑                      n            =            1                    ∞                                                    (              −                              1                2                            −              n              +              1              )                                      (              −                              1                2                            −              n              +              2              )                        ⋯                          (              −                              1                2                            )                                            n            !                                                (            −                          1              2                        )                    n                                                  =                  ∑                      n            =            0                    ∞                                                    (                                                          −              1                              /                            2                        n                                              )                                                            (            −                          1              2                        )                    n                −        1                                          =        (        1        −        1                  /                2                  )                      −            1                          /                        2                          −        1                                          =                  2                −        1.

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