If $f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$, find the required value We have
$$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$, then find the value of $\sum_{k=1}^{3} k f(k)$. 
Could someone give me slight hint as how to find $f(x)$ here. 
 A: $$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$
$$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{n^2}{n^2+x^2r^2}$$
$$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+x^2\cdot \frac{r^2}{n^2}}$$
$$= \lim_ {h \to 0} \text{h} \sum_{r=1}^{n} \frac{1}{1+x^2(rh)^2}$$
Then using Riemann summation, we get 
$$f(x)=\int_0^1 \frac{1}{1+x^2y^2}dy$$
$$=\frac{1}{x}\cdot \int_0^x \frac{d(xy)}{1+(xy)^2}$$
$$=\frac{\arctan x}{x}$$
So $$\boxed{f(x)=\frac{\arctan x}{x}}$$
Hence, $$\sum_{k=1}^{3} k f(k)=1f(1)+ 2f(2)+ 3f(3)=\arctan 1+\arctan 2+\arctan 3= \pi$$
Hope this helps you.
A: Since the terms of the sum are non-negative and monotonically decreasing in $r$, we can write
$$\int_1^{n+1} \frac{n}{n^2+x^2y^2}\,dy \le \sum_{r=1}^n\frac{n}{n^2+x^2r^2}\le \int_0^n \frac{n}{n^2+x^2y^2}\,dy  \tag 1$$
Evaluation of the integrals in $(1)$ is straightforward and reveals
$$\frac{\arctan\left(\frac{x}{1+(n+1)x^2/n^2}\right)}{x}\le \sum_{r=1}^n\frac{n}{n^2+x^2r^2}\le \frac{\arctan(x)}{x}$$
whereupon application of the squeeze theorem yields the coveted result
$$\lim_{n\to \infty}\sum_{r=1}^n\frac{n}{n^2+x^2r^2}=\frac{\arctan(x)}{x}$$
