Answer: (a) \(\displaystyle \sum_{r=1}^{n} \frac{n}{n^{2}+r^{2}}<\frac{\pi}{4}\).

(b) A lower bound is \(\displaystyle \sum_{r=1}^{n} \frac{n}{n^{2}+r^{2}}\ge \frac{\pi}{4}-\frac{1}{2n}\).

(c) \(\displaystyle \lim_{n\to\infty}\sum_{r=1}^{n}\frac{n}{n^{2}+r^{2}}=\frac{\pi}{4}\).

The curve is \(y=\frac{1}{x^2+1}\) on \(0\le x\le 1\). Since \(\frac{1}{x^2+1}\) is decreasing for \(x\ge 0\), right-end rectangles lie below the curve and left-end rectangles lie above the curve.

(a) Using \(n\) rectangles of width \(\frac{1}{n}\) with right endpoints \(x=\frac{r}{n}\), the total area is

\(\displaystyle \frac{1}{n}\sum_{r=1}^{n}\frac{1}{1+\left(\frac{r}{n}\right)^2}\;=\;\frac{1}{n}\sum_{r=1}^{n}\frac{n^2}{n^2+r^2}\;=\;\sum_{r=1}^{n}\frac{n}{n^2+r^2}.\)

These rectangles are below the curve, so

\(\displaystyle \sum_{r=1}^{n}\frac{n}{n^2+r^2}<\int_0^1\frac{1}{x^2+1}\,dx.\)

Now

\(\displaystyle \int_0^1\frac{1}{x^2+1}\,dx=[\tan^{-1}x]_0^1=\frac{\pi}{4}.\)

Hence

\(\displaystyle \sum_{r=1}^{n}\frac{n}{n^2+r^2}<\frac{\pi}{4}.\)

(b) Use left-end rectangles with the same width \(\frac{1}{n}\). Their total area is

\(\displaystyle \frac{1}{n}\sum_{r=0}^{n-1}\frac{1}{1+\left(\frac{r}{n}\right)^2}\;=\;\frac{1}{n}\sum_{r=0}^{n-1}\frac{n^2}{n^2+r^2}.\)

These rectangles lie above the curve, so

\(\displaystyle \frac{1}{n}\sum_{r=0}^{n-1}\frac{n^2}{n^2+r^2}\ge \int_0^1\frac{1}{x^2+1}\,dx=\frac{\pi}{4}.\)

Now rewrite the left-hand side:

\(\displaystyle \frac{1}{n}\sum_{r=0}^{n-1}\frac{n^2}{n^2+r^2}=\frac{1}{n}+\sum_{r=1}^{n-1}\frac{n}{n^2+r^2}.\)

Also,

\(\displaystyle \sum_{r=1}^{n-1}\frac{n}{n^2+r^2}=\sum_{r=1}^{n}\frac{n}{n^2+r^2}-\frac{n}{n^2+n^2}=\sum_{r=1}^{n}\frac{n}{n^2+r^2}-\frac{1}{2n}.\)

Therefore

\(\displaystyle \frac{1}{n}+\sum_{r=1}^{n}\frac{n}{n^2+r^2}-\frac{1}{2n}\ge \frac{\pi}{4}.\)

So

\(\displaystyle \sum_{r=1}^{n}\frac{n}{n^2+r^2}\ge \frac{\pi}{4}-\frac{1}{2n}.\)

(c) From parts (a) and (b),

\(\displaystyle \frac{\pi}{4}-\frac{1}{2n}\le \sum_{r=1}^{n}\frac{n}{n^2+r^2}<\frac{\pi}{4}.\)

As \(n\to\infty\), both bounds tend to \(\frac{\pi}{4}\). Hence

\(\displaystyle \lim_{n\to\infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\frac{\pi}{4}.\)