$$ \begin{aligned} \sum_{n=1}^{N} \frac{1}{n^{2}} & \leq 1+\sum_{n=2}^{N} \frac{1}{n^{2}-n} \\ &=1+\sum_{j=1}^{N-1} \frac{1}{n(n+1)} \end{aligned} $$
In this summation, the person changes the summation of expression
$\frac{1}{n^{2}-n}$ to $\frac{1}{n(n+1)}$. I understand that this works because he changed the limits of the expression too, but how did he come up with $\frac{1}{n(n+1)}$?
Could someone please take me through how he got that?
Thank you, and English isn't my first language -- Dutch is -- so apologies.
With the substitution $j=n-1$, it can be seen that $$\sum\limits_{n=2}^N\dfrac1{n(n-1)}=\sum\limits_{j=1}^{N-1}\dfrac1{(j+1)j}.$$