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 $\varepsilon > 0$.

I am confused. Are you taking the limit as $n$ tends to some value? If so, what value? — zz20s, Jan 04 '16 at 22:07
Your problem makes no sense. Is the limit for $n \to \infty$? or $n \to 0$? — David G. Stork, Jan 04 '16 at 22:08
Since we are talking natural numbers, I think the only sensible question to ask would be the limit for $n \to \infty$. — Bib-lost, Jan 04 '16 at 22:10
Since we are working with the natural numbers we could break this into two square roots... How would you define $\operatorname{frac}(\sqrt{n})$? — Brevan Ellefsen, Jan 04 '16 at 22:11
@djechlin Fractional part of $x$ is defined as $x$ minus the integer part of $x$. — N. S., Jan 04 '16 at 22:13

score 3 · Accepted Answer · edited Jan 04 '16 at 22:27

3

Hint:

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

edited Jan 04 '16 at 22:27

djechlin

answered Jan 04 '16 at 22:11

N. S.

score 2 · Answer 2 · answered Jan 04 '16 at 22:12

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.

answered Jan 04 '16 at 22:12

Brian Tung

2 Answers2