Find the limit of $\sum\limits_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$ 
$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$$

Note that $\forall x\ge 0, \sqrt{x}-1\le\sqrt{1+x}-1\le x$
Then
$$\sum_{k=1}^n\left(\sqrt{\frac{k}{n^2}}-1\right)\le S_n\le \sum_{k=1}^n\frac{k}{n^2}=\frac{1}{2}-\frac{1}{2n}$$
How do I evaluate the sum on the left?

For every $x$ in $(0,1)$, $\frac12x-x^2\leqslant\sqrt{1+x}-1\leqslant\frac12x$ hence
$$
\frac1{2n^2}T_n-\frac1{n^4}R_n\leqslant S_n\leqslant\frac1{2n^2}T_n,\qquad T_n=\sum_{k=1}^nk,\quad R_n=\sum_{k=1}^nk^2.
$$
One knows that $T_n=\frac12n(n+1)$ and $R_n\leqslant\sum\limits_{k=1}^nn^2=n^3$ hence
$$
\frac{n+1}{4n}-\frac1n\leqslant S_n\leqslant\frac{n+1}{4n},
$$
from which the limit of $S_n$ is easy to guess.
