 # I need to evaluate a series of a function that switches sign in the following way: <mtable colum Monserrat Sawyer 2022-05-24 Answered
I need to evaluate a series of a function that switches sign in the following way:
$\begin{array}{r}\sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\text{sgn}\left(n-k\right)}{\left(\left(2n+1\right)+B\text{sgn}\left(n-k\right)\right)-\left(2k+1\right)}\end{array}$
where $B\in \mathbb{R}$ and $n\in \mathbb{Z}$
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I would first shift the sum to $j=n-k$. Then your sum is
$\begin{array}{rc}& \sum _{k=-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\text{sgn}\left(n-k\right)}{\left(\left(2n+1\right)+B\text{sgn}\left(n-k\right)\right)-\left(2k+1\right)}\\ =& \sum _{j=-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\text{sgn}\left(j\right)}{2j+B\text{sgn}\left(j\right)}\\ =& \sum _{j=1}^{+\mathrm{\infty }}\frac{1}{2j+B}-\sum _{j=-1}^{-\mathrm{\infty }}\frac{1}{2j-B}\\ =& \sum _{j=1}^{+\mathrm{\infty }}\frac{1}{j+B/2}=-\frac{2}{B}-{\psi }^{\left(0\right)}\left(B/2\right),\end{array}$
where I replaced $j\to -j$ in the second sum in the second line to move to the third line.