Find $\lim_{n\to \infty}({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}})$

Note

$$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+n}}\le\lim_\limits{n\to \infty}\left({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}}\right)\le\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+1}}$$

Since $$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+n}}=\lim_\limits{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}}=1$$ and $$\lim_\limits{n\to \infty}\frac{n}{\sqrt{n^2+1}}=\lim_\limits{n\to \infty}\frac{1}{\sqrt{1+\frac{1}{n^2}}}=1$$

we have that the limit of the original is $1$ by the sandwich rule.