Consider the following sum: S ( n ) = ∑ k = 0 ∞ ( 2 k + n k ) 2 k + n 1 2 2 k ; n = 0 , 1 , 2 , 3 , . . .

Question

Consider the following sum:  S  (  n  )  =      ∑          k      =      0              ∞                                    (                                      2          k          +          n                k                              )                            2      k      +      n            1          2              2        k              ;  n  =  0  ,  1  ,  2  ,  3  ,  .  .  .

Karla Hull · Accepted Answer

The sum at hand is a hypergeometric series. Let      c    k    =      1          n      +      2      k                          (                            n        +        2        k            k                      )                  1          2              2        k              =            (      n      −      1      +      2      k      )      !              k      !      (      n      +      k      )      !            1          4      k      Indeed, the hypergeometric certificate is:            c              k        +        1                    c      k        =      1    4              (      n      +      2      k      )      (      n      +      2      k      +      1      )              (      n      +      1      +      k      )      (      k      +      1      )      Meaning that      ∑          k      =      0        ∞        c    k    =      c    0        ∑          k      =      0        ∞                                (          n                      /                    2          )                k                              (          n                      /                    2          +          1                      /                    2          )                k                    (      n      +      1              )        k                  1          k      !        =      1    n    ⋅              2        F    1        (          n      2        ,                  n        +        1            2        ;    n    +    1    ;    1    )  where   (  a      )    k   denotes Pochhammer symbol. Using Gauss&#039;s theorem, applicable for   ℜ  (  c  −  a  −  b  )  &amp;gt;  0              2        F          1            (    a    ,    b    ;    c    ;    1    )    =            Γ      (      c      )      Γ      (      c      −      a      −      b      )              Γ      (      c      −      a      )      Γ      (      c      −      b      )      we have      ∑          k      =      0        ∞        c    k    =      1    n              Γ      (      n      +      1      )      Γ              (                  1          2                )                    Γ              (                  n          2                +        1        )            Γ              (                              n            +            1                    2                )                                =                    duplication                  1    n                      2                  n                            π                  −          1                      /                    2                    Γ              (                              n            +            1                    2                )            Γ              (                  n          2                )            Γ              (                  1          2                )                    Γ              (                  n          2                )            Γ              (                              n            +            1                    2                )              =            2      n        n  Since   Γ  (  1      /    2  )  =      π

Consider the following sum: S ( n ) = <munderover> ∑ <mrow cl

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