Evaluation of Integral sums

Question

Hey I am supposed to evaluate: $$\lim_{n \to \propto }\sum_{i=1}^{n}\frac{i}{n^{2}+i^{2}}$$

What I did:

$$\lim_{n \to \propto }\sum_{i=1}^{n}\frac{i}{n^{2}+i^{2}}=\lim_{n \to \propto }\frac{1}{n^{2}}\sum_{i=1}^{n}\frac{i}{1+\left ( \frac{i}{n} \right )^{2}}$$

But I do not know, what to do next, or how to eliminate i so I can transfer it to integral.

Can anyone help me?

Answer 1

We have that

$$\sum_{i=1}^{n}\frac{i}{n^{2}+i^{2}}=\frac n{n^2}\sum_{i=1}^{n}\frac{\frac in}{1+\left(\frac in\right)^2}=\frac1n\sum_{i=1}^{n}\frac{\frac in}{1+\left(\frac in\right)^2}$$

then use Riemann sum.

Answer 2

$$\dfrac r{n^2+r^2}=\dfrac1n\cdot\dfrac{r/n}{1+(r/n)^2}$$

Now use The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$

Answer 3

Note that $$\lim_{n\to\infty}\sum_{i=1}^n\frac{i}{n^2+i^2}$$ $$=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\frac{\frac{i}{n}}{1+\frac{i^2}{n^2}}$$ $$=\int_0^1\frac{x}{1+x^2}dx$$

I think you can proceed hereon....