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. 

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
$$\lim_{n\to \infty}\sum_{r=1}^n\frac{n}{n^2+x^2r^2}=\frac{\arctan(x)}{x}$$
