the question is
Consider the sequence $a_n = \{({{n*\sqrt{2}}})\}, n ≥ 1. $ where {} means the fractional part of $a$. Show that the sequence an contains a subsequence that converges to 0
here's my attempt. I think about using Continued fraction since we know $$ \sqrt{2} = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \dots}}, $$
if I can pick certain numbers to prove the pattern to be monotonic with strictly decreasing, then it's done? but I am stuck here and don't know how to deal with it. Any suggestion will be helpful, thanks!
or maybe I can do the transform such that $f_0=0, f_1=\sqrt2-1= \frac{1}{\sqrt2+1}$, I found that $f_5 =5\sqrt2-7= \frac{1}{\sqrt5+7} $. (i use $f$ stands for fractional part) it may work if there's more pattern like this, but it's probably a dead end.
I don't know if this post will help, but I can't really understand what's the first answer doing there.
i notice that using Bolzano–Weierstrass can prove it to be bounded but how to show it's necessarily leading to 0?
So , $$a_n=\{n\sqrt2\}=n\sqrt2-\lfloor n\sqrt2\rfloor$$
and we know $\forall n\in\mathbb N$ $$0\leq n\sqrt2-\lfloor n\sqrt2\rfloor<1\implies0\leq a_n<1$$
Hence the sequence is bounded .
Now according to Bolzano Weierstrass theorem which states that Every bounded sequence has a convergent subsequence , proves that your sequence also has a convergent subsequence