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

I don't understand how can I start the above problem because Riemann sums and integrals are not applicable here.
It seems that we can solve it using  the Squeeze theorem, but I don't understand
which inequality I have to use. Help required, thanks.

Note that $\sqrt{1+k/n^2} -1 = \frac{k}{2n^2} + O\left(\left( \frac{k}{n^2} \right)^2\right)$ by the Maclaurin series of $\sqrt{1+x}$. We use the notation $O(f(n))$ for some term thats bounded for sufficiently large $n$ by $C |f(n)|$ for some constant $C>0$ . 
So, $\sum_{k=1}^n \left( \sqrt{1+k/n^2} -1 \right)  = \sum_{k=1}^n \left( \frac{k}{2n^2} + O\left(\left( \frac{k}{n^2} \right)^2\right) \right) = \frac{n(n+1)}{2} \frac{1}{2n^2} + O\left(n\left( \frac{k}{n^2} \right)^2 \right) = \frac{n(n+1)}{2} \frac{1}{2n^2} + O(1/n)$ where the first equalityfollows from maclaurin series, second equality follows from evaluating the sum of the first term, and the sum containing $n$ terms, and the third equality follows from $k \leq n$. 
Now, take the limit as $n \to \infty$ to get $\frac{1}{4}$. 
