We have this sequence
$$\left\{\{\sqrt n\}\right\}_{n=1}^\infty\;,\;\;\{\sqrt n\}:=\;\text{the fractional part of}\;\;\sqrt n$$
The exercise is to prove it doesn't have a limit, and we get several hints:
(1) Show zero is a partial limit .- this is easy as the subsequence $\;\left\{\{\sqrt{n^2}\}\right\}_{n=1}^\infty\;$ is identical with zero
(2) Prove that for any $\;M\in\Bbb N\;$ there exists $\;n>M\;$ such that $\;\{\sqrt n\}\ge\frac12\;$ .
This is the REAL problem: it's easy to show that $\;a_{N^2+1}=\left\{\sqrt{N^2+1}\right\}>0\;$ . Then the following misterious, for me, hint is given: if $\;a_{N^2+1}\ge\frac12\;$ then choose $\;n=N^2+1\;$ and we're done (this is clear to me), otherwise $\;0<a_{N^2+1}<\frac12\;$ . Show there exists $\;k\in\Bbb N\;$ s.t. $\;a_{k^2(N^2+1)}\ge\frac12\;$ . Now
$$\;a_{k^2(N^2+1)}=\left\{\sqrt{k^2(N^2+1)}\right\}=\left\{k\sqrt{N^2+1}\right\}$$
This seems to hint towards taking integer multiples of $\;\sqrt{N^2+1}\;$ , and this seem very close to Dirichlet's Approximation theorem (with the pigeohole principle), yet I can't see how to show directly what's been asked without relying on Dirichlet's theorem, which is forbidden.
Just consider $a_{M^2+M+1}$. Since $M^2+M+1<(M+1)^2$, we just have to prove: $$\sqrt{M^2+M+1}-M\geq\frac{1}{2},$$ that is equivalent to: $$M^2+M+1\geq \left(M+\frac{1}{2}\right)^2,$$ that is equivalent to: $$ 1\geq \frac{1}{4}.$$