In the book of Difference Equations by Peterson, it is given that
$\Delta_n (\sum_{k=a}^{n-1} y(k)) = y(n),$ where $\Delta_n y(n) = y(n+1) - y(n)$.
So, if we can find $\sum y(n)$ (where $\sum$ in there stands for the inverse operation of $\Delta$, and it is called summand of $y(n)$), we can find the partial sum of $\sum_{k=a}^{n-1} y(k)$ - it is $\sum y(n)$.
Question:
What is the summand of $y(n) = \frac{1}{n^2}$ ?
Just for those who are not familiar with this notation:
$$\sum (\Delta_n y(n)) = y(n) \quad \& \quad \Delta_n(\sum y(n)) = y(n).$$