Use of digamma function to evaluate some finite series

Question

I need to evaluate a series of a function that switches sign in the following way: \begin{eqnarray} \sum_{k=-\infty}^{+\infty}\frac{\text{sgn}(n-k)}{((2n+1)+B\text{sgn}(n-k)) -(2k+1)} \end{eqnarray} where $B\in\mathbb{R}$ and $n\in\mathbb{Z}$. Due to this sign change, I am considering splitting the sum into three terms $\sum_{k\leq 0},\sum_{1\leq k\leq n-1}$ and $\sum_{k\geq n+1}$, since at $k=n$ the summand is zero. Also note that $\sum_{k\leq 0},\sum_{1\leq k\leq n-1}$ do carry the same sign; however my reason to do this is to use the following expression for the digamma function: \begin{eqnarray}\label{digam} \psi^{(0)}(z)=(-1)\sum_{k=0}^{+\infty}\frac{1}{k+z} \end{eqnarray} I can evaluate the first sum with $\sum_{k\leq 0}$ easily using the above formula; but I struggle to see how to compute the other terms, since their limits are different from $0$ and $+\infty$. How can I evaluate finite sums $\sum_{k=n_{1}}^{n_{2}}$ with $n_{2}>n_{1}$ in the spirit of using the above definition for $\psi^{(0)}(z)$?

Answer 1

I would first shift the sum to $j=n-k$. Then your sum is $$ \begin{eqnarray} & \sum_{k=-\infty}^{+\infty}\frac{\text{sgn}(n-k)}{((2n+1)+B\text{sgn}(n-k)) -(2k+1)}\\ =&\sum_{j=-\infty}^{+\infty}\frac{\text{sgn}(j)}{2j+B\text{sgn}(j)}\\ =&\sum_{j=1}^{+\infty}\frac{1}{2j+B}-\sum_{j=-1}^{-\infty}\frac{1}{2j-B}\\ =&\sum_{j=1}^{+\infty}\frac{1}{j+B/2}=-\frac{2}{B}-\psi^{(0)}(B/2), \end{eqnarray}$$ where I replaced $j\rightarrow-j$ in the second sum in the second line to move to the third line.

If you want to evaluate without shifting like I did, then you would split into $k<n, k=n$ and $k>n$. You were splitting into a region involving $k=0$, which wasn't relevant and I think this was confusing you.