Answer: (a) \(\displaystyle \sum_{r=1}^{N} \frac{1}{r^{2}} \gt \frac{3}{2}-\frac{1}{N}+\frac{1}{N^{2}}\).
(b) \(\displaystyle \sum_{r=1}^{N} \frac{1}{r^{2}} \lt \frac{7}{4}-\frac{1}{N}\).
(c) Hence, letting \(N\to\infty\),
\(\displaystyle \frac{3}{2}\lt \sum_{r=1}^{\infty} \frac{1}{r^{2}} \lt \frac{7}{4}\).
The curve is \(y=x^{-2}=\frac{1}{x^2}\), which is decreasing for \(x\ge 2\).
Also,
\(\displaystyle \int_2^N x^{-2}\,dx=\left[-x^{-1}\right]_2^N=\frac12-\frac1N.\)
(a)
Using unit-width rectangles from \(x=2\) to \(x=N\) with heights taken at the left-hand endpoints, the total area of the rectangles is
\(\displaystyle \frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{(N-1)^2}=\sum_{r=2}^{N-1}\frac1{r^2}.\)
Since the function is decreasing, these rectangles lie above the curve, so
\(\displaystyle \sum_{r=2}^{N-1}\frac1{r^2} \gt \int_2^N x^{-2}\,dx=\frac12-\frac1N.\)
Now
\(\displaystyle \sum_{r=1}^N \frac1{r^2}=1+\sum_{r=2}^{N-1}\frac1{r^2}+\frac1{N^2}.\)
Therefore
\(\displaystyle \sum_{r=1}^N \frac1{r^2} \gt 1+\left(\frac12-\frac1N\right)+\frac1{N^2}=\frac32-\frac1N+\frac1{N^2}.\)
(b)
For an upper bound, use unit-width rectangles with heights taken at the right-hand endpoints. Their total area is
\(\displaystyle \frac{1}{3^2}+\frac{1}{4^2}+\cdots+\frac{1}{N^2}=\sum_{r=3}^{N}\frac1{r^2}.\)
These rectangles lie below the curve, so
\(\displaystyle \sum_{r=3}^{N}\frac1{r^2} \lt \int_2^N x^{-2}\,dx=\frac12-\frac1N.\)
Now
\(\displaystyle \sum_{r=1}^N \frac1{r^2}=1+\frac14+\sum_{r=3}^{N}\frac1{r^2}.\)
Hence
\(\displaystyle \sum_{r=1}^N \frac1{r^2} \lt 1+\frac14+\left(\frac12-\frac1N\right)=\frac74-\frac1N.\)
(c)
From parts (a) and (b),
\(\displaystyle \frac32-\frac1N+\frac1{N^2} \lt \sum_{r=1}^N \frac1{r^2} \lt \frac74-\frac1N.\)
As \(N\to\infty\), the partial sums approach \(\displaystyle \sum_{r=1}^{\infty}\frac1{r^2}\), while
\(\displaystyle \frac32-\frac1N+\frac1{N^2}\to\frac32, \qquad \frac74-\frac1N\to\frac74.\)
So the infinite sum satisfies
\(\displaystyle \frac32 \lt \sum_{r=1}^{\infty}\frac1{r^2} \lt \frac74.\)
