I have the following sequence and limit:

$$\left(a_n\right)=\left(n-\sqrt{n}\left\lfloor{\sqrt{n}}\right\rfloor\right)$$ $$\lim_{n\to\infty}{n\left(\sqrt{n^2+2\ell}-n\right)}=\ell,$$

where $\ell$ is natural (i.e., $\ell \in \mathbb{N}$).

I have to prove that any natural number is a partial limit of the sequence $\left(a_n\right)$.

As I understand, I have to find a sequence of indexes that include $\ell$, that I can use to prove the claim. I've found a sequence of indexes that leads to a solution, but it is not a set of natural numbers. I can't get over this problem and build a sequence of natural indexes.

I thought of this 4 days, and I can't solve it, so I'm asking for help. This was my best solution, but the indexes are not natural numbers:

$$ \left(a_n\right)=\left(n-\sqrt{n}\left\lfloor{\sqrt{n}}\right\rfloor\right)= \sqrt{n}\left(\sqrt{n}-\left\lfloor{\sqrt{n}}\right\rfloor\right) $$ $$ \left(n_k\right)={\left(n+\frac {\ell}{n}\right)}^2$$ $$a_{n_k}=\sqrt{{\left(n+\frac {\ell}{n}\right)}^2}\left(\sqrt{{\left(n+\frac {\ell}{n}\right)}^2}-\left\lfloor{\sqrt{{\left(n+\frac {\ell}{n}\right)}^2}}\right\rfloor\right)=$$ $$=\left(n+\frac {\ell}{n}\right)\left(\sqrt{n^2+2\ell+\frac {\ell^2}{n^2}}-\lfloor{n+\frac {\ell}{n}}\rfloor\right)$$ $$\lim_{n\to\infty}{\left(n+\frac {\ell}{n}\right)\left(\sqrt{n^2+2\ell+\frac {l^2}{n^2}}-\lfloor{n+\frac {\ell}{n}}\rfloor\right)}=\lim_{n\to\infty}{n\left(\sqrt{n^2+2\ell}-n\right)}=\ell$$

  • $\begingroup$ Why do you have to prove this? $\endgroup$ – barak manos Jul 26 '16 at 9:35
  • $\begingroup$ As an exercise in the univercity $\endgroup$ – Oleg Jul 26 '16 at 9:37
  • 2
    $\begingroup$ In the current form of your question, you're not asking for help, you're asking that someone else will do it for you. Asking for help means showing what you've tried and where you got stuck. $\endgroup$ – barak manos Jul 26 '16 at 9:39
  • 1
    $\begingroup$ This was my best solution $\endgroup$ – Oleg Jul 26 '16 at 9:41
  • 1
    $\begingroup$ Then post it as part of the question. $\endgroup$ – barak manos Jul 26 '16 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.