Prove $\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$

Question

Desmos suggests that$$\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\psi(x)-\psi\left(\frac x2\right)-\ln2$$Where $\psi$ is the digamma function. I can write the LHS as $$\frac12\left(\psi\left(\frac{x+1}2\right)-\psi\left(\frac x2\right)\right)=\sum_{k=0}^\infty\frac{(-1)^k}{k+x}$$But the series of $\ln2$ makes turning the RHS into the LHS more difficult. I get that the LHS is $$\psi(x)-\psi\left(\frac x2\right)-\ln2=\sum_{k=0}^\infty\left(\frac1{k+\frac x2}-\frac1{k+x}+\frac{(-1)^k}k\right)$$

Answer 1

Desmos does not suggest that:

Quite the discrepancy. Rather: $$\tag{$\star$}\frac{1}{2}\left(\psi\left(\frac{x+1}{2}\right)+\psi\left(\frac{x}{2}\right)\right)=\psi(x)-\ln2$$With no superfluous "$-\psi(x/2)$" term.

How do I know $(\star)$? It follows easily from logarithmically differentiating the Gauss multiplication formula, or more simply the Legendre duplication formula (which is a special case). We have: $$\sum_{k=0}^{n-1}\psi^{(m)}\left(z+\frac{k}{n}\right)=n^{m+1}\cdot\psi^{(m)}(nz),\,m\ge1\\\sum_{k=0}^{n-1}\psi\left(z+\frac{k}{n}\right)=n(\psi(nz)-\ln n)\\\psi(z)+\psi\left(z+\frac{1}{2}\right)=2\psi(2z)-2\ln2$$

Answer 2

If you mean you want to prove this $$ \begin{aligned} & \sum_{k=0}^{2N} \left(\frac1{k+\frac{x}{2}} - \frac1{k+x}\right) \\ =&\sum_{k=0}^{2N} \frac2{2k+x} - \sum_{k=0}^{N} \frac1{2k+x} - \sum_{k=0}^{N-1} \frac1{2k+1+x}\\ =&\sum_{k=N+1}^{2N} \frac2{2k+x} + \left(\sum_{k=0}^{N} \frac1{2k+x}-\sum_{k=0}^{N-1} \frac1{2k+1+x}\right) \\ =&\sum_{k=1}^{N} \left(\frac2{2N+2k+x}-\frac1{N+k}\right) + \sum_{k=1}^{N} \frac1{N+k} + \left(\sum_{k=0}^{N} \frac1{2k+x}-\sum_{k=0}^{N-1} \frac1{2k+1+x}\right) \\ =&-\sum_{k=1}^{N} \frac{x}{(2N+2k+x)(N+k)} + \sum_{k=1}^{N} \frac1{N+k} + \left(\sum_{k=0}^{N} \frac1{2k+x}-\sum_{k=0}^{N-1} \frac1{2k+1+x}\right) \end{aligned} $$ Now, let $N\to\infty$ $$ \sum_{k=1}^{N} \frac{x}{(2N+2k+x)(N+k)} = \frac1{N} \sum_{k=1}^{N} \frac{x}{(2+\frac{2k+x}{N})(1+\frac{k}{N})}\frac1{N} \to 0 $$ $$ \sum_{k=1}^{N} \frac1{N+k} \to \ln2 $$ $$ \sum_{k=0}^{N} \frac1{2k+x}-\sum_{k=0}^{N-1} \frac1{2k+1+x} = \sum_{k=0}^{2N} \frac{(-1)^k}{k+x} \to \sum_{k=0}^{\infty} \frac{(-1)^k}{k+x} $$ Then you have your conclusion. (I edited for some mistake)