Prove: $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)$ Prove: $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}=\frac{2}{3}(2\sqrt{2}-1)$
What method to use in order to find the closed form of summation $\sum\limits_{k=1}^{n}\sqrt{1+\frac{k}{n}}$?
 A: The sums
$$\frac{1}{n}\sum_{k = 1}^n \sqrt{1 + \frac{k}{n}}$$
are Riemann sums of the continuous function $f(x) = \sqrt{1 + x}$ over the interval $[0,1]$. So your sequence converges to 
$$\int_0^1 \sqrt{1 + x}\, dx = \int_1^2 \sqrt{x}\, dx,$$
which I leave to you to evaluate.
A: METHOD 1:  Use the Euler-Maclaurin Formula 
While the easier way is perhaps to recognize the limit as a Riemann sum, I thought it might be instructive to present a way forward that used the Euler-Maclaurin Formula.  To that end, we write
$$\begin{align}
\sum_{k=1}^n\sqrt{1+\frac kn}&=\int_0^n \sqrt{1+\frac xn}\,dx+\frac12 \left(\sqrt{2}-1\right)+\frac1{24\,n}\left(\frac{\sqrt{2}}{2}-1\right)+O\left(\frac{1}{n^2}\right)\\\\
&=\frac{2n}{3}\left(2^{3/2}-1\right)+\frac12 \left(\sqrt{2}-1\right)+\frac1{24\,n}\left(\frac{\sqrt{2}}{2}-1\right)+O\left(\frac{1}{n^2}\right)
\end{align}$$
Dividing by $n$ and taking the limit produces the expected result!

METHOD 2:  Use the Squeeze Theorem
Here we use the squeeze theorem. Note that the sum of interest is bounded as
$$\int_0^n\sqrt{1+\frac xn}\,dx\le \sum_{k=1}^n\sqrt{1+\frac kn}\le \int_1^{n+1} \sqrt{1+\frac kn}\,dx$$
Carrying out the integrals yields
$$\frac{2n}{3}\left(2^{3/2}-1\right)\le \sum_{k=1}^n\sqrt{1+\frac kn}\le \frac{2n}{3}\left(2^{3/2}\left(1+\frac1{2n}\right)-\left(1+\frac1{n}\right)\right)$$
whereupon dividing by $n$ and invoking the Squeeze Theorem yields
$$\lim_{n\to \infty}\frac1n \sum_{k=1}^n\sqrt{1+\frac kn}=\frac23 \left(2^{3/2}-1\right)$$
