Let $\{a_n\}_{n=0}^{\infty}$ be a sequence of positive real numbers such that $\limsup_n \frac{1}{n}\log a_n=-\infty$. Then $$ f(x)=\sum_{n\geq 0}a_n x^n $$ converges absolutely for all $x$. Under what conditions on $\{a_n\}$ will we have $f(x)\geq 0$ for all $x\leq 0$?

Unanswered on MSE

  • 4
    $\begingroup$ I think this is unlikely to have meaningful answers in this form; the question seems too general. $\endgroup$ – Christian Remling Jun 28 '15 at 5:26
  • $\begingroup$ Is this really an open problem? That would surprise me somehow... $\endgroup$ – pre-kidney Jun 28 '15 at 14:03
  • $\begingroup$ may the sequence $\{a_n\}_{n=0}^{\infty}$ should be decreasing and converge $\endgroup$ – zeraoulia rafik Jun 28 '15 at 22:14
  • $\begingroup$ The sequence $\{a_n\}_{n=0}^{\infty}$ should be decreasing and converge vers $0$ as :lim$suplog (a_{n})=-\infty $ , $n \to \infty$ hence you can use leibnez theorem for (Alternating series) $\endgroup$ – zeraoulia rafik Jun 28 '15 at 22:29

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.