Deriving $\sum\limits_{i=1}^n \frac{1}{i^2} = \frac{π^2}{6}-ψ^{(1)}(n+1)$. How is the partial sums formula of the $$\sum\limits_{i=1}^n \frac{1}{i^2} = \frac{π^2}{6}-ψ^{(1)}(n+1)$$
derived?

It can be shown with the Hurwitz Zeta function (see especially formulas  (1) and (2))
$$\sum\limits_{i=1}^n \frac{1}{i^2} 
= \sum\limits_{i=1}^{\infty} \frac{1}{i^2} - \sum\limits_{i=n+1}^{\infty} \frac{1}{i^2} \\
= \sum\limits_{i=1}^{\infty} \frac{1}{i^2} - \sum\limits_{i=0}^{\infty} \frac{1}{(i+n+1)^2} \\
= \zeta(2)-\zeta(2,n+1)\\
=  \frac{\pi^2}{6}-ψ^{(1)}(n+1)$$
and the well known value $\zeta(2) = \frac{\pi^2}{6}$ for the Riemann Zeta function.
