Evaluate $\lim_{n \to \infty} \sum_{i=0}^{2n} \frac{i}{n^2 + i^2}$

Question

This is my work so far:

$$\lim_{n \to \infty} \sum_{i=0}^{2n} \frac{i}{n^2 + i^2} = \lim_{n \to \infty} \sum_{i=0}^{2n} \frac{\frac{i}{n^2}}{1 + \frac{i^2}{n^2}} = \lim_{n \to \infty} \frac{1}{n}\sum_{i=0}^{2n} \frac{\frac{i}{n}}{1 + (\frac{i}{n})^2} $$

I am not sure where to go form here, by testing out larger values of $n$ I get that this limit seems to approach the decimal $\cong 0.804$, but I'm not sure how to get this.

Answer 1

Here's how to find a function which has this sum as a Riemann sum on a certain interval:

Set $\;x_i=\dfrac in,\enspace y_i=1+x_i=\dfrac{n+i}n,\enspace\Delta x=\Delta y=\dfrac1n,\;$ and rewrite the sum: \begin{align} \frac1n\sum_{i=0}^{2n} \frac{\frac{i}{n}}{1 + (\frac{i}{n})^2}&=\frac1n\sum_{i=1}^{n} \frac{\frac{i}{n}}{1 + (\frac{i}{n})^2}+\frac1n\sum_{i=1}^{n} \frac{\frac{n+i}{n}}{1 + (\frac{n+i}{n})^2}\\[1ex] &=\Delta x \sum_{i=0}^{n} \frac{x_i}{1 + x_i^2}+\Delta y \sum_{i=0}^{n} \frac{y_i}{1 + y_i^2} \\[1ex] &\to \int_0^1\frac x{1+x^2}\,\mathrm d x +\int_1^2\frac y{1+y^2}\,\mathrm d y \\[1ex] &= \int_0^2\frac x{1+x^2}\,\mathrm d x =\frac12\,\ln(1+x^2)\biggr|_0^2 =\color{red}{\frac12\ln 5}. \end{align}