If $q_n$ be an enumeration of rational numbers, for which $x$ the following sequence converges?


I guess that for no $x$ the sequence converges. I tried to consider a subsequence of $q_n$ converging to $x$ such that $e^{-n_k^2|x-q_{n_k}|}$ does not converge to zero.

  • 2
    $\begingroup$ Integrating over $\mathbb R$ we see that the series converges for almost all $x$. $\endgroup$ – Kavi Rama Murthy Jan 7 '19 at 23:53
  • $\begingroup$ What kind of answer are you looking for? "For what $x$" is very unclear. Do you have to show that there can only be countably many $x$ where it diverges, for example? $\endgroup$ – A. Pongrácz Jan 8 '19 at 0:42

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.