Prove that limit of the fractional part of $\sqrt{n^2+n}$ is $\frac{1}{2}$

Question

Prove that $$\operatorname{frac}(\sqrt{n^2 + n}) \to \frac{1}{2}$$ ($n \in \mathbb{N}$, $\operatorname{frac}$ is fractional part of number)

I think I should use just definition of limit and find $N$ for all $\varepsilon > 0$.

Answer 1

Hint:

$$\left(n+\frac{1}{2}\right)^2 \geq n^2+n \geq \left(n+\frac{1}{2} -\frac{1}{n}\right)^2$$

Answer 2

Hint. Consider that

$$ \left(\sqrt{n^2+n}+n\right)\left(\sqrt{n^2+n}-n\right) = (n^2+n)-n^2 = n $$

Note that $2n < \sqrt{n^2+n}+n < 2n+1$ and squeeze.