Prove that $1.49<\sum_{k=1}^{99}\frac{1}{k^2}<1.99$ 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.
 A: Hint: Telescoping and $$1+\sum_{k=2}^{99}\frac{1}{k\;(k+1)}<\sum_{k=1}^{99}\frac{1}{k^2}<1+\sum_{k=2}^{99}\frac{1}{k\;(k-1)}$$
A: For the lower bound, just compute the first few terms:
$$
\sum_{k=1}^{6}\frac{1}{k^2} \approx 1.4913 > 1.49
$$
The exact value is
$$
\sum_{k=1}^{6}\frac{1}{k^2} = \frac{5369}{3600} = 1.4913\bar8
$$
A: Following the creative telescoping approach I used to solve another question,
$$\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}\leq\sum_{n\geq m}\frac{1}{n^2}\leq\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}\tag{1}$$
hence:
$$\frac{1}{m}-\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}\leq\sum_{n> m}\frac{1}{n^2}\leq\frac{1}{m}-\frac{1}{2m^2}+\frac{1}{6m^3}\tag{2}$$
and by setting $m=100$ we have:
$$ 0.009950166663\ldots\leq\frac{\pi^2}{6}-H_{100}^{(2)}\leq 0.009950166666\ldots\tag{3} $$
so:

$$ \sum_{n=1}^{99}\frac{1}{n^2}=\color{red}{1.6348839}+E,\qquad |E|\leq \color{red}{10^{-9}}. \tag{4}$$

A: This approach as inspired by the Riemann sum. Basically, comparing the area of rectangular strips of unit width and height $\frac 1{k^2}$ with the area under the curve $y=\frac 1{k^2}$.
$$\begin{align}
1+\int_2^{100}\frac 1{k^2}dk<&\sum_{k=1}^{99}\frac 1{k^2}<1+\int_1^{99}\frac 1{k^2}dk\\
1.49<&\sum_{k=1}^{99}\frac 1{k^2}<1.99
\end{align}$$
