Sum to infinity of trignometry inverse: ∑ r = 1 ∞ arctan ⁡ ( 4 r 2 + 3 )

Question

Sum to infinity of trignometry inverse:       ∑          r      =      1        ∞    arctan  ⁡      (          4                        r          2                +        3              )

hoffwnbu · Accepted Answer

∑                      r            =            1                    ∞                arctan        ⁡                  (                      4                                          r                2                            +              3                                )                                    =                  ∑                      r            =            1                    ∞                          [          arctan          ⁡                      (                          2                              r                −                1                                      )                    −          arctan          ⁡                      (                          2                              r                +                1                                      )                    ]                                                  =                  lim                      n            →            ∞                                    [                      ∑                          r              =              1                        n                    arctan          ⁡                      (                          2                              r                −                1                                      )                    −                      ∑                          r              =              3                                      n              +              2                                arctan          ⁡                      (                          2                              r                −                1                                      )                    ]                                                  =                  lim                      n            →            ∞                                    [                      π            2                    +          arctan          ⁡          (          2          )          −          arctan          ⁡                      (                          2              n                        )                    −          arctan          ⁡                      (                          2                              n                +                1                                      )                    ]                                                  =                  π          2                +        arctan        ⁡        (        2        )

Richardtb · Accepted Answer

One may observe that summing      u          r      +      1        −      u          r      −      1      may be simplified as a telescoping sum:      ∑          r      =      1        N        (          u              r        +        1              −          u              r        −        1              )    =      ∑          r      =      1        N        (          u              r        +        1              −          u              r              )    +      ∑          r      =      1        N        (          u              r              −          u              r        −        1              )    =      u          N      +      1        +      u    N    −      u    1    −      u    0    .From the identity  arctan  ⁡      (          4                        r          2                +        3              )    =  arctan  ⁡      (                  r        +        1            2        )    −  arctan  ⁡      (                  r        −        1            2        )  by telescoping you then obtain                              ∑                      r            =            1                    N                arctan        ⁡                  (                      4                                          r                2                            +              3                                )                                    =        arctan        ⁡                  (                                    N              +              1                        2                    )                +        arctan        ⁡                  (                      N            2                    )                                                                −        arctan        ⁡                  (                      1            2                    )                −        arctan        ⁡                  (                                    1              −              1                        2                    )                ,            giving, as   N  →  ∞ ,      ∑          r      =      1        ∞    arctan  ⁡      (          4                        r          2                +        3              )    =  2  arctan  ⁡      (    ∞    )    −  arctan  ⁡      (          1      2        )    =      π    2    +  arctan  ⁡  2.

Sum to infinity of trignometry inverse: <munderover> ∑ <mrow class="M

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