# Closed form solution to the Digamma recurrence relation?

The difference between two digamma function can be written using the following recurrence relation:

$\psi(n+z) - \psi(z) = \sum_{i=0}^{n} \frac{1}{i + z}$

My question is, is there a closed form solution to this recurrence function? and how can i get?

I appologize my question was not clear. I want a 'function' to compute the right hand side efficiently $\sum_{i=0}^{n-1} \frac{1}{i + z}$, without the need to do linear increment if possible and also without using the digamma function. I do not want to use the digamma function because it is expensive. I used this recurrence relation for the difference of the two digamma function in the first place to avoid evaluating $\psi(x)$.

• Thank you. I'm going to delete my answer. – Olivier Oloa Aug 31 '16 at 14:43
• I don't see how it's computationally expensive to do this unless you want to do it an enormous amount of times and over a large range of $z$ and $n$. In that case why not just compute and spline $\psi(z)$ over a large range in $z$ (possibly in conjuncting with using analytical expressions for asymptotical behavior) and then use this spline to compute $\psi(n+z) - \psi(z)$ whenever needed. – Winther Aug 31 '16 at 15:04