Assistance in finding $\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n}$

Question

I am trying to prove $$\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n} = 0$$ given $n>0$.

But I'm having difficulties dealing with the floor function. I tried splitting apart the limit like so:

\begin{align*} \lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n} &= \frac{\lim_{n\to\infty} (n-\lfloor \sqrt n \rfloor^2)}{\lim_{n\to\infty} n} \\ &= \frac{\lim_{n\to\infty} n}{\lim_{n\to\infty} n} - \frac{\lim_{n\to\infty} \lfloor \sqrt n \rfloor^2}{\lim_{n\to\infty} n} \end{align*}

Trivially $\lim_{n \to \infty} \lfloor \sqrt n \rfloor^2 = \infty$, but it seems to grow somewhat slower than $\lim_{n \to \infty} n$ and so I am not sure if it is correct to conclude that $$\frac{\lim_{n\to\infty} \lfloor \sqrt n \rfloor^2}{\lim_{n\to\infty} n} = 1.$$

Answer 1

Using $$x-1<\lfloor x \rfloor \le x$$ $$\lim_{n\to\infty} \frac{n-(\sqrt n)^2}{n}\le\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n}\le\lim_{n\to\infty} \frac{n-(\sqrt n-1)^2}{n}$$ $$0\le\lim_{n\to\infty} \frac{n-\lfloor \sqrt n \rfloor^2}{n}\le\lim_{n\to\infty} \frac{2\sqrt n-1}{n}=0$$

Answer 2

For very large values of $n$, the floor of $\sqrt{n}$ is almost indistinguishable (in the relative sense) from $\sqrt{n}$ itself; the absolute difference is at most $1$, so the relative difference is tending to zero.

To make this precise, note that we have the inequalities

$$\sqrt{n} - 1 \le \lfloor \sqrt{n} \rfloor \le \sqrt{n}$$ Thus upon squaring,

$$n - 2\sqrt{n} + 1 \le \lfloor \sqrt{n}\rfloor^2 \le n$$

After some algebraic work, you should be able to reduce the problem to computing $\lim_{n \to \infty} \sqrt{n} / n$.

Answer 3

Another possibility:

$$\frac{n-\lfloor \sqrt n \rfloor ^2}{n} \leq \frac{\lceil \sqrt{n} \rceil^2 - \lfloor \sqrt n \rfloor ^2}{n} = \frac{(\lceil \sqrt{n} \rceil - \lfloor \sqrt n \rfloor)(\lceil \sqrt{n} \rceil + \lfloor \sqrt n \rfloor) }{n} \leq \frac{2 \lceil \sqrt{n} \rceil}{n} \to 0.$$

Answer 4

The lowest that $\lfloor\sqrt n\rfloor$ can be is $\sqrt n - 1$. Hence we have

$$n - \lfloor\sqrt n\rfloor^2 > n - (n - 2\sqrt n +1) = 2\sqrt n - 1$$

So the above divided by $n$ tends to zero.