Can someone help me in proving the following: π 2 = ∑ l = 0 ∞ ( − 1 ) l 2 l + 1 ( P 2 l ( x ) + sgn ( x ) ⋅ P 2 l + 1 ( x ) ) ,

Question

Can someone help me in proving the following:      π    2    =      ∑          l      =      0        ∞              (      −      1              )        l                    2      l      +      1        (      P          2      l        (  x  )  +  sgn  (  x  )  ⋅      P          2      l      +      1        (  x  )  )  ,

Tianna Deleon · Accepted Answer

Assume   x  ∈      R  ,       |    x      |    &amp;lt;  1. Then        |        P    n    (  x  )      |    =  O  (      n          −      1              /            2        ) as   n  →  ∞, hence  F  (  x  ,  z  )  :=      ∑          n      =      0        ∞        P    n    (  x  )            z              n        +        1                    n      +      1        ,    G  (  x  ,  z  )  :=      ∑          n      =      1        ∞        P    n    (  x  )            z      n        n  converge absolutely and uniformly in   {  z  ∈      C    :      |    z      |    ⩽  1  }, and clearly                    F        (        x        )        :=                  ∑                      n            =            0                    ∞                                      (            −            1                          )              n                                            2            n            +            1                                    P                      2            n                          (        x        )                            =                  1                      2            i                                                (                          F        (        x        ,        i        )        −        F        (        x        ,        −        i        )                              )                          ,                            G        (        x        )        :=                  ∑                      n            =            0                    ∞                                      (            −            1                          )              n                                            2            n            +            1                                    P                      2            n            +            1                          (        x        )                            =                  1                      2            i                                                (                          G        (        x        ,        i        )        −        G        (        x        ,        −        i        )                              )                          .            Both   F  (  x  ,  z  ) and   G  (  x  ,  z  ) can be computed using the generating function  P  (  x  ,  t  )  :=      ∑          n      =      0        ∞        P    n    (  x  )      t    n    =      1          1      −      2      x      t      +              t        2            (here   t  ∈      C  ): for       |    z      |    &amp;lt;  1 we then have                    F        (        x        ,        z        )                            =                  ∫          0          z                P        (        x        ,        t        )                d        t                                  =        log        ⁡                              z            −            x            +                          1              −              2              x              z              +                              z                2                                                          1            −            x                          ,                            G        (        x        ,        z        )                            =                  ∫          0          z                                      P            (            x            ,            t            )            −            1                    t                        d        t                                  =        log        ⁡                  2                      1            −            x            z            +                          1              −              2              x              z              +                              z                2                                                    ,            and for       |    z      |    =  1 we may take radial limits, by Abel&#039;s theorem. This gives, for       x    ⩾    0    F  (  x  ,  ±  i  )  =  log  ⁡            ±      i      +              x                    1      +              x              ,    G  (  x  ,  ±  i  )  =  log  ⁡      2          (      1      +              x            )      (      1      ∓      i              x            )        ,then   F  (  x  )  =  arccot  ⁡      x   and   G  (  x  )  =  arctan  ⁡      x  , hence   F  (  x  )  +  G  (  x  )  =  π      /    2The case       |    x      |    &amp;lt;  1 follows by parity, and the cases   x  =  ±  1 are easy to check separately.

Can someone help me in proving the following: π 2 </mfrac> = <mu

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