Prove that for every $n \in \mathbb{N}$ $\sum\limits_{k=2}^{n}{\frac{1}{k^2}}<1$

Let $ n\in\mathbb{N} : $

Notice that for any integer $ k\geq 2 $, we have $ k^{2}\geq k\left(k-1\right) $, and thus : $$ \sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\sum_{k=2}^{n}{\frac{1}{k\left(k-1\right)}}=\sum_{k=2}^{n}{\left(\frac{1}{k-1}-\frac{1}{k}\right)}=1-\frac{1}{n}<1 $$

Strengthen the induction hypothesis

Hint: Prove by induction that

$$ \sum_{k=2}^n \frac{1}{k^2 } < 1 - \frac{1}{n}.$$