First analysis course and first question here, so excuse me if I'm missing something.

The question is: what is the supremum and infimum of the set of points $$ { \sqrt n - floor(\sqrt n)} $$ with n natural.

It is trivial to prove that 0 and 1 are lower and upper bounds, but I have had great difficulty showing they are the infimum and supremum.

Thank you!


1 Answer 1


Let $f(n) = \sqrt n - \lfloor \sqrt n \rfloor$. Consider $$\lim_{n \to \infty} f(n^2 + 1)$$ and $$\lim_{n \to \infty} f(n^2 - 1)$$

It's easy to see that the first limit is $0$ and the second limit is $1$, thus demonstrating the value of the infinum and the supremum.

To see the first limit, note that $$f(n^2 + 1) = \sqrt {n^2 + 1} - n = \frac {(\sqrt {n^2 + 1} - n)(\sqrt {n^2 + 1} + n)}{\sqrt {n^2 + 1} + n} = \frac {1}{\sqrt {n^2 + 1} + n}$$

Taking limits throughout, we easily see that the right-hand side approaches $0$ so the left hand side also approaches $0$. Using a similar proof we can easily see that $\sqrt{n^2-1}$ approaches $n$ for large $n$, so $f(n^2-1)$ approaches $1$.

  • $\begingroup$ Thank you, those two are very elegant solutions! However, I don't see how I would actually go about proving those limits. The floor operator makes it hard to find a relationship between epsilon and n, and I have not been able to come up with sequences appropriate for the squeeze theorem. Could you help me with that as well? $\endgroup$ Feb 3, 2019 at 8:38
  • $\begingroup$ I've revised the answer to clarify the method of proof. $\endgroup$ Feb 4, 2019 at 4:30

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