Re-Expressing the Digamma

The following formulas may be inserted into the formula given by @Lucian

$$H_n=\sum_{k=1}^n\frac{1}{k} $$

$$ H_n-log\left(n\right)=1-\sum_{k=1}^{\infty}\left(\sum_{i=nk+1}^{n(k+1)}\frac{1}{i}-\frac{1}{k+1}\right) $$ https://math.stackexchange.com/a/1602945/134791

$$ \gamma= \sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right)=\sum_{k=1}^\infty \left(\frac{2}{k}-\sum_{j=k(k-1)+1}^{k(k+1)} \frac{1}{j}\right) $$

https://math.stackexchange.com/a/1591256/134791

$$\psi_0\left(n+1\right)=H_n-\gamma$$

$$=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^\infty \left(\frac{2}{k}-\sum_{j=k(k-1)+1}^{k(k+1)} \frac{1}{j}\right)$$

In the complex plane outside negative integers (http://people.math.sfu.ca/~cbm/aands/page_259.htm) $$\psi(z+1)=-\gamma+\sum_{n=1}^\infty\frac{z}{n(n+z)}=-\gamma+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+z}\right)$$ $$=-\sum_{n=1}^\infty \left(\frac{2}{n}-\sum_{j=n(n-1)+1}^{n(n+1)} \frac{1}{j}\right)+\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+z}\right)$$

Finally, $$\psi(z+1)=\sum_{n=1}^\infty \left(-\frac{1}{n}-\frac{1}{n+z} +\sum_{j=n(n-1)+1}^{n(n+1)}\frac{1}{j} \right), z \neq -1,-2,-3,...$$