A function $\psi(z)$ satisfies the functional equation $\psi(z+1)=\frac{1}{z}+\psi(z)$. Prove that $\psi(\frac{1}{2})=\psi(1)-2\ln2$. I was given the following question to solve:

Given a function $\psi(z)$ satisfies the following functional equation:
$$\psi(z+1)=\frac{1}{z}+\psi(z)$$
Prove that $\psi(\frac{1}{2})=\psi(1)-2\ln2$.

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 $\psi(z)$. Help would be much appreciated.
 A: 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:
$$\sum_{k=1}^{+\infty}\left(\frac{2}{2k+1}-\frac{1}{k}\right)=2\int_{0}^{1}\sum_{k\geq 1}\left(x^{2k}-x^{2k-1}\right)\,dx = -2\int_{0}^{1}\frac{x\,dx}{1+x}=-2(1-\log 2).$$
