How would you compute the following series? I'm interested in some easy approaches that would allow me to work it out. ∑ n = 1 ∞ 4 n n ! ( 2 n ) !

Question

How would you compute the following series? I&#039;m interested in some easy approaches that would allow me to work it out.      ∑          n      =      1              ∞                          4        n            n      !              (      2      n      )      !

nuvolor8 · Accepted Answer

It suffices to calculate the sum with the summation index running from 0, not from 1. Then                              ∑                      n            =            0                                ∞                                                              4              n                        n            !                                (            2            n            )            !                                              =                  ∑                      n            =            0                                ∞                                                4            n                                (            2            n            )            !                                    ∫                      0                                ∞                                    t          n                          e                      −            t                                  d        t                                          =                  ∫                      0                                ∞                                    ∑                      n            =            0                                ∞                                                              4              n                                      t              n                                            (            2            n            )            !                                    e                      −            t                                  d        t                                          =                  ∫                      0                                ∞                          cosh        ⁡        (        2                  t                )                  e                      −            t                                  d        t                                          =                  ∫                      0                                ∞                          2        u        cosh        ⁡        (        2        u        )                  e                      −                          u              2                                              d        u                (        t        =                  u          2                )                                          =                              [            −            cosh            ⁡            (            2            u            )                          e                              −                                  u                  2                                                      ]                                0                                ∞                          +        2                  ∫                      0                                ∞                          sinh        ⁡        (        2        u        )                  e                      −                          u              2                                              d        u                                          =        1        +        2                  ∫                      0                                ∞                          sinh        ⁡        (        2        u        )                  e                      −                          u              2                                              d        u                                          =        1        +                  ∫                      0                                ∞                                    (                      e                          2              u                                −                      e                          −              2              u                                )                          e                      −                          u              2                                              d        u                                          =        1        +        e                  ∫                      0                                ∞                                    (                      e                          −              (              u              −              1                              )                2                                              −                      e                          −              (              u              +              1                              )                2                                              )                        d        u        .            Now it is clear how to express the last integrals in terms of the error function, yielding      ∑          n      =      1              ∞                          4        n            n      !              (      2      n      )      !        =  e      π        e    r    f    (  1  )  .

kixEffinsoj · Accepted Answer

Here is a generalization. Consider the generating function  f  (  x  )  =      ∑          n      ≥      1                          x                  2          n                    n      !              (      2      n      )      !        .It&#039;s an easy check that 2 is in the radius of convergence, so that we need to compute   f  (  2  ) to solve your problem. However, it&#039;s easy to see that      f    ″    (  x  )  =  1  +      ∑          n      ≥      1                          x                  2          n                    (      n      +      1      )      !              (      2      n      )      !      Moreover,  x      f    ′    (  x  )  =  2      ∑          n      ≥      1                          x                  2          n                    n      !              (      2      n      )      !        ⋅  nWe get the second-order ODE            x              f        ′            (      x      )        2    +  f  (  x  )  =      f    ″    (  x  )  −  1  ,  f  (  0  )  =  0  ,      f    ′    (  0  )  =  0.And the solution  f  (  x  )  =      1    2    ⋅      e                  x        2                    /            4        ⋅  x      π        e    r    f    (  x      /    2  )  .Just plug in   x  =  2 to get your answer,   e      π        e    r    f    (  1  )

How would you compute the following series? I'm interested in some easy approaches that would allow

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