I am reading through the Wikipedia article on the digamma function where the digamma function is compared to the Harmonic Numbers defined as $H_n = \sum\limits_{k=1}^{n} \dfrac{1}{k}$.
The first equation is well explained and clear to me:
$$\psi(n) = H_{n-1} - \gamma$$
It's the last equation which is not clear to me from the argument:
$$\psi\left(n+\frac{1}{2}\right) = -\gamma -2\ln 2 + \sum\limits_{k=1}^{n}\frac{2}{2k-1}$$
How does this equation follow from the explanation that came before?
Could someone help me to understand why it is true?
$\Gamma(s+1)=s\Gamma(s)$ gives $\psi(s+1)=\psi(s)+1/s$, $\psi(n+1/2) =\psi(1/2)+ \sum_{k=1}^n \frac1{k-1/2}$ so it is just saying that $\psi(1/2)=-\gamma -2\log 2$ which follows from the explicit formula $\psi(z)=-\gamma-\sum_n \frac1{z+n}-\frac1{n+1}$ and the expansion of $\log (1+x),x\to 1$