In a question at CrossValidated we have a user asking for derivation of the "reflection formula" for the digamma function. Using integers $0 \leqslant x \leqslant n$ the formula of interest is:
$$\psi (n+1) - \psi(n-x+1) = \sum_{i=1}^x \frac{1}{n-x+i}.$$
Can one of you learned people supply a derivation of this formula?
That's a very weird way to write it, since it's more usual to treat the $x$ as a continuous variable and $n$ as a discrete variable (usually an integer), but we can derive the required relation as:
$$\begin{eqnarray}\psi(z) & = & \frac{\Gamma'(z)}{\Gamma(z)} \\ \psi(z + 1) & = & \frac{\Gamma'(z + 1)}{\Gamma(z + 1)} \\ & = & \frac{(z \Gamma(z))'}{z \Gamma(z)} \\ & = & \frac{\Gamma(z) + z \Gamma'(z)}{z \Gamma(z)} \\ & = & \psi(z) + \frac{1}{z} \end{eqnarray}$$
That gives us a recurrence relation for $\psi$. We can then repeatedly apply it thus:
$$\begin{eqnarray}\psi(z + 1) & = & \psi(z) + \frac{1}{z} \\ & = & \psi(z - 1) + \frac{1}{z - 1} + \frac{1}{z} \\ & = & \psi(z - 2) + \frac{1}{z - 2} + \frac{1}{z - 1} + \frac{1}{z} \\ & \cdots \\ & = & \psi(z - x + 1) + \frac{1}{z - x + 1} + \ldots + \frac{1}{z} \\ & = & \psi(z - x + 1) + \sum_{i = 1}^{x} \frac{1}{z - x + i} \end{eqnarray}$$
And just replace $z$ with $n$ to get your expression. I would more normally write it as $\psi(z + n) = \psi(z) + \sum_{i = 0}^{n - 1} \frac{1}{z + i}$ but the two are equivalent.