I need to evaluate a series of a function that switches sign in the following way: ∑ k = − ∞ + ∞ sgn ( n − k ) ( ( 2 n + 1 ) + B sgn ( n − k ) ) − ( 2 k + 1 ) where B ∈ R and n ∈ Z

Question

I need to evaluate a series of a function that switches sign in the following way:                              ∑                      k            =            −            ∞                                +            ∞                                                sgn            (            n            −            k            )                                (            (            2            n            +            1            )            +            B            sgn            (            n            −            k            )            )            −            (            2            k            +            1            )                              where   B  ∈      R   and   n  ∈      Z

Harper Heath · Accepted Answer

I would first shift the sum to   j  =  n  −  k. Then your sum is                                            ∑                      k            =            −            ∞                                +            ∞                                                sgn            (            n            −            k            )                                (            (            2            n            +            1            )            +            B            sgn            (            n            −            k            )            )            −            (            2            k            +            1            )                                              =                                      ∑                      j            =            −            ∞                                +            ∞                                                sgn            (            j            )                                2            j            +            B            sgn            (            j            )                                              =                                      ∑                      j            =            1                                +            ∞                                    1                      2            j            +            B                          −                  ∑                      j            =            −            1                                −            ∞                                    1                      2            j            −            B                                              =                                      ∑                      j            =            1                                +            ∞                                    1                      j            +            B                          /                        2                          =        −                  2          B                −                  ψ                      (            0            )                          (        B                  /                2        )        ,            where I replaced   j  →  −  j in the second sum in the second line to move to the third line.

I need to evaluate a series of a function that switches sign in the following way: <mtable colum

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