If $f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$, find the required value

Question

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.

Answer 1

$$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$ $$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{n^2}{n^2+x^2r^2}$$ $$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+x^2\cdot \frac{r^2}{n^2}}$$ $$= \lim_ {h \to 0} \text{h} \sum_{r=1}^{n} \frac{1}{1+x^2(rh)^2}$$ 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.

Answer 2

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

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