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I was given the following question to solve:

"Given a function $\psi_0(z)$ satisfies the following functional equation:

$\psi_0(z+1)=\frac{1}{z}+\psi_0(z)$

Prove that $\psi_0(\frac{1}{2})=\psi_0(1)-2\ln2$."

I'm not sure how to go about approaching this question; letting $z=0$ isn't possible and I'm not clear as to whether I should find a way to define $\psi_0(z)$. Help would be much appreciated.

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If $$\psi(z+1) = \frac{1}{z}+\psi(z) \tag{1}$$ by induction it follows that: $$\forall n\in\mathbb{N}^+,\quad \psi(n)= H_{n-1}+\psi(1). \tag{2}$$ On the other hand, $(1)$ implies that $\psi(z+1)-\psi(z)$ is positive, but decreasing to zero, as $z\to+\infty$. This gives that $\psi(z)$ is a concave function and: $$ \lim_{z\to +\infty}\left(\psi\left(x+\frac{1}{2}\right)-\psi(x)\right)=0.\tag{3}$$ However: $$ \psi\left(n+\frac{1}{2}\right)=\frac{2}{2n-1}+\psi\left((n-1)+\frac{1}{2}\right) = \sum_{k=0}^{n-1}\frac{2}{2k+1}+\psi\left(\frac{1}{2}\right)\tag{4},$$ hence by subtracting $(2)$ from $(4)$ we get: $$ \psi\left(n+\frac{1}{2}\right)-\psi(n) = 2+\sum_{k=1}^{n-1}\left(\frac{2}{2k+1}-\frac{1}{k}\right)+\psi\left(\frac{1}{2}\right)-\psi(1)\tag{5}$$ and the claim follows from considering the limit as $n\to +\infty$, since:

$$\sum_{k=1}^{+\infty}\left(\frac{2}{2k+1}-\frac{1}{k}\right)=2\int_{0}^{1}\sum_{k\geq 1}\left(x^{2k}-x^{2k-1}\right)\,dx = -2\int_{0}^{1}\frac{x\,dx}{1+x}=-2(1-\log 2).$$

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