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

Question

$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$$

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?

Answer 1

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.

Answer 2

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.