Compute the sum: ∑ n = 1 ∞ 1 ⋅ 3 ⋯ ( 2 n − 1 ) 2 ⋅ 4 ⋯ ( 2 n ) ⋅ ( 2 n + 1 )

Question

Compute the sum:      ∑          n      =      1              ∞                          1        ⋅        3        ⋯        (        2        n        −        1        )                    2        ⋅        4        ⋯        (        2        n        )        ⋅        (        2        n        +        1        )

percolarse2rzd · Accepted Answer

Starting with the power series derived using the binomial theorem,  (  1  −  x      )          −      1              /            2        =  1  +            1      2        x  +            1      2                  3      2            x    2        /    2  !  +            1      2                  3      2                  5      2            x    3        /    3  !  +  ⋯  +                    1        ⋅        2        ⋅        3        ⋯        (        2        n        −        1        )                    2        ⋅        4        ⋅        6        ⋯        2        n                  x    n    +  ⋯and integrating, we get the series for      sin          −      1        ⁡  (  x  )  =      ∫    0    x    (  1  −      t    2        )          −      1              /            2            d    t  =      ∑          n      =      0        ∞                      1        ⋅        2        ⋅        3        ⋯        (        2        n        −        1        )                    2        ⋅        4        ⋅        6        ⋯        2        n                        x              2        n        +        1                    2      n      +      1      Setting   x  =  1, we get      ∑          n      =      1        ∞                      1        ⋅        2        ⋅        3        ⋯        (        2        n        −        1        )                    2        ⋅        4        ⋅        6        ⋯        2        n                  1          2      n      +      1        =      sin          −      1        ⁡  (  1  )  −  1  =      π    2    −  1

arbixerwoxottdrp1l · Accepted Answer

We have      ∑          n      =      1              ∞                          1        ⋅        3        ⋯        (        2        n        −        1        )                    2        ⋅        4        ⋯        (        2        n        )        ⋅        (        2        n        +        1        )              =      ∑          n      =      1        ∞                                (                                      2          n                n                              )                          1                  4        n            (      2      n      +      1      )        .Now notice that      ∫    0          1              /            2        d  x    2      x          2      n        =      1                  4        n            (      2      n      +      1      )        .Thus, we need only compute the sum                    2                  ∑                      n            =            1                    ∞                                                                    (                                                                    2                n                            n                                                      )                                                              x                      2            n                                              =                    2                  ∑                      n            =            0                    ∞                                                                    (                                                                    2                n                            n                                                      )                                                              x                      2            n                          −        2                                          =                              2                      1            −            4                          x              2                                      −        2            and integrate.

Compute the sum: <munderover> ∑ <mrow class="MJX-TeXAtom-ORD"> n

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