Prove that every natural number is a limit of a subsequence

Question

I need to prove that every natural number is a partial limit (limit of a subsequence) to the sequence ${{a}_{n}}=n-{{\left\lfloor \sqrt{n} \right\rfloor }^{2}}$ . I already found that 0 is a partial limit. I was given a hint that for every natural l exists $n<\sqrt{{{n}^{2}}+l}<n+1$, I’ve proven this equation. Now what I tried to do is try and use it by: ${{\left\lfloor \sqrt{n} \right\rfloor }^{2}}<{{\left\lfloor \sqrt{n} \right\rfloor }^{2}}+l<{{\left( \left\lfloor \sqrt{n} \right\rfloor +1 \right)}^{2}}$so that $\left| n-{{\left\lfloor \sqrt{n} \right\rfloor }^{2}}-l \right|=\left| n-({{\left\lfloor \sqrt{n} \right\rfloor }^{2}}+l) \right|$ would somehow help me achieve < $\varepsilon $ But this is where I was stuck for a few hours.

Answer 1

HINT: Take the hint a little further: $n<\sqrt{n^2+\ell}<n+1$ if and only if $0\le\ell<2n+1$. In particular, if $\ell\in\Bbb N$, $n\in\Bbb N$, and $n>\frac12(\ell-1)$, then $n<\sqrt{n^2+\ell}<n+1$. That gives you a choice of infinitely many values of $n$ making $n<\sqrt{n^2+\ell}<n+1$.

• Now suppose that $n<\sqrt{n^2+\ell}<n+1$; what is $a_{n^2+\ell}$?

Answer 2

One way to see this is by noticing that $a_n$ is linear on all range of the form $[n^2,(n+1)^2]$, this is basically what the hint is telling you.

So on each interval of the form $[n^2,(n+1)^2]$, $a_n$ start at $0$ and increases linearly until it values $(n+1)^2-n^2-1$. So for all $n$ s.t. $(n+1)^2-n^2-1>l$, $a_n$ will periodically values $l$ at least once in each of the subsequent intervals.

More formally :

Let $l\in \mathbb{N}$, $l$ is a partial limit if $\forall M \in \mathbb{N}, \exists n$ s.t. $n>M$ and $n-\lfloor\sqrt{n}\rfloor^2=l$

Let $M \in \mathbb{N}$, and $n=(M+l)^2+l$, then $(M+l)^2+l < (M+l)^2+2(M+l)+l^2 = (M+l+1)^2$ $\Rightarrow (M+l)<\lfloor\sqrt{(M+l)^2+l}\rfloor<(M+l+1)$

Thus $\lfloor\sqrt{(M+l)^2+l}\rfloor=(M+l)$

and $a_{(M+l)^2+l}=(M+l)^2+l-(M+l)^2=l$

So $l$ is a partial limit of $a_n$.