Evaluate the following limit: $=\lim\limits_{n\to \infty} \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $ I am studying MIT OpenCourseware 18.01 Single Variable Calculus on my own and am stuck on a final exam question.
Evaluate the following limit: 
$$\lim\limits_{n\to \infty} \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}}  \left(\frac{2}{n}\right)$$
$$=\lim\limits_{n\to \infty}  \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $$ 
We can do this using the Riemann sum, which states that if the interval [a,b] is divided into $n$ equal pieces of length, where $\Delta x = \frac{b-a}{n}$, then the sum of all the areas of the rectangle is $ \sum_{i = 1}^{n} f(x_{i-1}) \Delta x $. Also, in the limit as $n$ goes to infinity, the Riemann sum approaches the value of the definite integral:
$$\lim\limits_{n\to \infty} \sum_{i = 1}^{n} f(x_{i-1}) \Delta x  =\int_a^b f(x)\,dx$$
In this case,  $\Delta x = \frac{b-a}{n} = \frac{2}{n}$, and therefore, $b-a = 2$. Also $f(x_0) = \sqrt{1+\frac{2}{n}}$, $f(x_1) = \sqrt{1+\frac{4}{n}}$, $f(x_2) = \sqrt{1+\frac{6}{n}}$, and so on and so forth until we reach $n$.
How do we convert $$\lim\limits_{n\to \infty}  \left(\frac{2}{n}\right) \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}} $$ to a definite integral?

Just for your curiosity (without Riemann sum).
Assuming that you know about generalized harmonic numbers, you could approximate quite well the partial sums since 
$$ S_n= \frac{2}{n} \sum_{i = 1}^{n} \sqrt{1+\frac{2i}{n}}=2 \sqrt{2}\,\frac{ H_{\frac{3
   n}{2}}^{\left(-\frac{1}{2}\right)}-H_{\frac{n}{2}}^{\left(-\frac{1}{2}\right)}
  }{n^{3/2}}$$
Now, using the asymptotics
$$H_p^{\left(-\frac{1}{2}\right)}=\frac{2 p^{3/2}}{3}+\frac{p^{1/2}}{2}+\zeta
   \left(-\frac{1}{2}\right)+\frac{1}{24p^{1/2}}-\frac{1}{1920p^{5/2}}+O\left(
   \frac{1}{p^{9/2}}\right)$$
$$S_n=\left(2 \sqrt{3}-\frac{2}{3}\right)+\frac{\sqrt{3}-1}{n}+\frac{\sqrt{3}-3}{18
   n^2}+\frac{27-\sqrt{3}}{3240 n^4}+O\left(\frac{1}{n^6}\right)$$ which, for sure, shows the limit and also how it is approached.
Moreover, it is a quite good approximation. For example, computing $S_5\approx 2.9410395$ while the above truncated formula would give $\frac{4459499 \sqrt{3}-1768473}{2025000}\approx 2.9410399$.
