How do I evaluate t$\lim_{n\to\infty} \sum_{i=1}^n \left[\sqrt{1+ \frac{2i}{n}}\right]\frac{2}{n}$? (From MIT OCW 18.01 sc final Q7(a)) This was one of the questions on the final for MIT's 18.01:
$$\lim_{n\to\infty} \sum_{i=1}^n \left[\sqrt{1+ \frac{2i}{n}}\right]\frac{2}{n}$$
The answer converts it to an integral, but I'm not sure how they made that logical step. Is this using L'Hopital's rule somehow?
 A: The technique used for computing this series is a Riemann-Summation type argument: The interval $I=(0,1)$ is partitioned in $n$ subintervals of length $\color{blue}{\Delta x_i=\frac{1}{n}}$. Then, for each of these, the point $\color{red}{x_i^\ast=\frac{i}{n}}$ is chosen in each of these intervals. If the Riemann summation converges we then know that
$$\sum_{i}f(x_i^\ast)\Delta x_i \to \int_If(x)\,dx$$
thus in this particular case, as we see the expression
$$\sum_{i}\sqrt{1+2\cdot \color{red}{\frac{i}{n}}}\cdot \color{blue}{\frac{1}{n}}=\sum_{i}\underbrace{\sqrt{1+2\cdot \color{red}{x_i^\ast}}}_{f(x_i)}\cdot \color{blue}{\Delta x_i}\to\int_I\sqrt{1+2x}\,dx$$
from here it's easy to conclude.
A: An (upper) Riemann sum is given by $$\int_{x=a}^b f(x) \, dx = \lim_{n \to \infty}  \sum_{i=1}^n f\left(a + \frac{b-a}{n} i\right) \frac{b-a}{n}.$$  So with the choice $a = 1$, $b = 3$, $f(x) = \sqrt{x}$, we obtain $$\lim_{n \to \infty} \sum_{i=1}^n \sqrt{1 + \frac{2i}{n}} \frac{2}{n} = \int_{x=1}^3 \sqrt{x} \, dx = \frac{2}{3}(3^{3/2} - 1).$$
