summation - Prove that $1.49

It can be proven by induction that $$\sum_{k=1}^{n}\frac{1}{k^2}\leq2-\frac{1}{n}$$

From here, we can easily acquire the upper bound of the sum $$\sum_{k=1}^{99}\frac{1}{k^2}$$ letting $n=100$.

However, I am not quite sure about the lower bound. The standard method of constructing lower rectangles of unit width on the curve $y=\frac{1}{x^2}$ yields a lower bound of $0.99$, which isn't tight enough unfortunately. What could I do to achieve a sharper bound?

I know that as $n\rightarrow \infty$, the sum converges to $\frac{\pi^2}{6}$, but I feel that proving that result, just for this inequality, is a bit much.