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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?

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    $\begingroup$ No. I think they are using Riemann Summation. Are you familiar with such argumentations? $\endgroup$ – b00n heT Jul 4 at 8:36
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    $\begingroup$ Are $[]$ supposed to refer to normal brackets? $\endgroup$ – Tavish Jul 4 at 8:36
  • $\begingroup$ Yes I think the square bracktes are normal brackets and yes i am familiar with Riemann summation $\endgroup$ – doctopus Jul 4 at 8:39
  • $\begingroup$ 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. $\endgroup$ – b00n heT Jul 4 at 8:40
  • $\begingroup$ Yes I think I get the gist of it.. it's still rusty though. A full answer would be fantastic. $\endgroup$ – doctopus Jul 4 at 8:47
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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.

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

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