<s>[INST] 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?
 [/INST] We have $$S_n=\sum_{1,n} \sqrt{1+k/n^2}-1=\sum_{1,n}\frac{k/n^2}{\sqrt{1+k/n^2}+1}$$
hence

$$\frac{n(n+1)}{2n^2 (\sqrt{1+1/n}+1)}=\sum_{1,n}\frac{k/n^2}{\sqrt{1+n/n^2}+1}\leq S_n\leq \sum_{1,n}\frac{k/n^2}{\sqrt{1}+1}=\frac{n(n+1)}{4n^2}$$

and so $\lim S_n =\frac{1}{4}$ by squeezing.
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