# 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?

• No. I think they are using Riemann Summation. Are you familiar with such argumentations? – b00n heT Jul 4 at 8:36
• Are $[]$ supposed to refer to normal brackets? – Tavish Jul 4 at 8:36
• Yes I think the square bracktes are normal brackets and yes i am familiar with Riemann summation – doctopus Jul 4 at 8:39
• Then you should recognize the Riemann Summation: $\frac{i}{n}$ takes the role of $x$ whilst $\frac{1}{n}$ "becomes" $dx$. I can expand it to a full answer if it helpful for you. – b00n heT Jul 4 at 8:40
• Yes I think I get the gist of it.. it's still rusty though. A full answer would be fantastic. – doctopus Jul 4 at 8:47

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.
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 $$\lim_{n \to \infty} \sum_{i=1}^n \sqrt{1 + \frac{2i}{n}} \frac{2}{n} = \int_{x=1}^3 \sqrt{x} \, dx = \frac{2}{3}(3^{3/2} - 1).$$