What does this series converge to? ∑ n = 0 ∞ ( γ 2 ) 2 n ( 2 n k 2 + n )

Question

What does this series converge to?      ∑          n      =      0              ∞        (      γ    2        )          2      n                          (                            2        n                              k          2                +        n                            )

Perman7z · Accepted Answer

Since   k  ∈  2      N   we have   m  =  k      /    2  ∈      N  . Now consider  S  (  x  ,  m  )  =      ∑          n      =      0              ∞                                    (                                      2          n                          m          +          n                                      )                          x    n    .The binomial coefficient is equal to zero when   n  &amp;lt;  m so  S  (  x  ,  m  )  =      ∑          n      =      m              ∞                                    (                                      2          n                          m          +          n                                      )                          x    n    =      x    m        ∑          n      =      0              ∞                                    (                                      2          n          +          2          m                          n          +          2          m                                      )                          x    n  Let   (  s      )    n    =  Γ  (  s  +  n  )      /    Γ  (  s  ) be the Pochhammer symbol. We have by the gamma duplication formula and definition of the binomial coefficient:                                                                        (                                                                    2                n                +                2                m                                            n                +                2                m                                                                    )                                                                        =                              Γ            (            2            n            +            2            m            +            1            )                                Γ            (            n            +            2            m            +            1            )            n            !                                                            =                              2                          2              n              +              2              m                                            π                                                Γ            (            m            +            1                          /                        2            +            n            )            Γ            (            m            +            1            +            n            )                                Γ            (            n            +            2            m            +            1            )            n            !                                                            =                              2                          2              m                                            π                                                Γ            (            m            +            1                          /                        2            +            n            )            Γ            (            m            +            1            +            n            )                                Γ            (            2            m            +            1            +            n            )                                                4            n                                n            !                                                            =                                                                                                  2                                          2                      m                                                                            π                                                                                        Γ                    (                    m                    +                    1                                          /                                        2                    )                    Γ                    (                    m                    +                    1                    )                                                        Γ                    (                    2                    m                    +                    1                    )                                                              ⏟                                            =            1                                                Γ            (            m            +            1                          /                        2            +            n            )            Γ            (            m            +            1            +            n            )            Γ            (            2            m            +            1            )                                Γ            (            m            +            1                          /                        2            )            Γ            (            m            +            1            )            Γ            (            2            m            +            1            +            n            )                                                4            n                                n            !                                                            =                              (            m            +            1                          /                        2                          )              n                        (            m            +            1                          )              n                                            (            2            m            +            1                          )              n                                                            4            n                                n            !                              Substituting this result back into S we have  S  (  x  ,  m  )  =      x    m        ∑          n      =      0              ∞                  (      m      +      1              /            2              )        n            (      m      +      1              )        n                    (      2      m      +      1              )        n                                4        x            n              n      !        .Finally, we substitute   x  =      γ    2        /    4,   m  =  k      /    2, and identify the series S as a hypergeometric function to get  S  (      γ    2        /    4  ,  k      /    2  )  =            (              γ        2            )        k                    2              F      1            (                                                      k              +              1                        2                    ,                      k            2                    +          1                          k          +          1                      ;          γ      2        )    .According to this hypergeometric reduction formula, we then simplify our solution to get the final form                    ∑                  n          =          0                          ∞                                                                (                                                          2              n                                      k                              /                            2              +              n                                                          )                                                            γ                      2            n                                    2                      2            n                              =                        γ          k                                      1            −                          γ              2                                (          1          +                      1            −                          γ              2                                            )            k                              .

What does this series converge to? <munderover> ∑ <mrow class="MJX-TeXA

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