Just as the topic ask how to evaluate ∑ k = 1 ∞ ( 18 ) [ ( k − 1 ) ! ] 2 ( 2 k ) ! .

Question

Just as the topic ask how to evaluate      ∑          k      =      1              ∞                  (      18      )      [      (      k      −      1      )      !              ]        2                    (      2      k      )      !        .

gaiageoucm5p · Accepted Answer

Notice that            (      k      −      1      )              !        2                    (      2      k      )      !        =            Γ      (      k      )      Γ      (      k      )              Γ      (      2      k      +      1      )        =      1          2      k        B  (  k  ,  k  )  =      1          2      k            ∫    0    1        x          k      −      1        (  1  −  x      )          k      −      1            d    xThus                              ∑                      k            =            1                    ∞                                      (            k            −            1            )                          !              2                                            (            2            k            )            !                                              =                              ∫          0          1                          (                      ∑                          k              =              1                        ∞                                1                          2              k                                            x                          k              −              1                                (          1          −          x                      )                          k              −              1                                )                          d                x        =                  ∫          0          1                                      −            ln            ⁡            (            1            −            x            +                          x              2                        )                                2            x            (            1            −            x            )                                    d                x                                          =                    −        2                  ∫                      −            1                          /                        2                                1                          /                        2                                                ln            ⁡            (            3                          /                        4            +                          u              2                        )                                1            −            4                          u              2                                                d                u        =        −        2                  ∫          0          1                                      ln            ⁡            (            (            3            +                          u              2                        )                          /                        4            )                                1            −                          u              2                                                d                u        =                              π            2                    18

mravinjakag · Accepted Answer

In order to fill in on the last equality, define  f  (  t  )  =  −  2      ∫    0    1              ln      ⁡              (        1        −        t        (        1        −                  u          2                )        )                    1      −              u        2                  d    uClearly   f  (  0  )  =  0, and we are interested in computing   f      (          1      4        )        f    ′    (  t  )  =  2      ∫    0    1                      d            u              1      +      t      (      1      −              u        2            )                          =                              u          =                                                    1                −                t                            t                                tan          ⁡          (          ϕ          )                          ∫    0          arcsin      ⁡      (              t            )                  2              d            ϕ              t      (      1      −      t      )        =            2      arcsin      ⁡      (              t            )              t      (      1      −      t      )        =    2            d                      d            t            arcsin    2    ⁡  (      t    )Thus  f      (          1      4        )    =      ∫    0                  1        4              2            d                      d            t            arcsin    2    ⁡  (      t    )  =  2      arcsin    2    ⁡      (          1      2        )    =            π      2        18

Just as the topic ask how to evaluate <munderover> ∑ <mrow class="MJX-T

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