1
$\begingroup$

I would like to know if the following assertion is true: Let consider a real decreasing sequence $(t_n)$ of positive numbers with limit zero, if the series $\sum\limits_{n=1}^\infty(t_n)^a$ is divergent for all real $a$ in $[0,1[$ then the series $\sum\limits_{n=1}^\infty t_n$ is also divergent? thanks in advance

$\endgroup$

closed as off-topic by Greg Martin, Jan-Christoph Schlage-Puchta, Lucia, fedja, Willie Wong Aug 31 '18 at 17:54

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Greg Martin, Lucia, fedja
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ For some basic information about typing mathematical expressions on Stack Exchange sites, see: How does one type mathematical formulas on this site? To help you get started, I have edited your post - I hope this is at least approximately close to what you intended, but if not please edit your post further. It was a bit unclear whether you want to ask about sequence or about series - in case it helps you, you can write sum as \sum $\sum$ or \sum_{n=1}^\infty $\sum_{n=1}^\infty$. (To get math rendered, it has to be included between dollar signs.) $\endgroup$ – Martin Sleziak Aug 27 '18 at 9:55
  • $\begingroup$ I mean " the series \sum tn^a is divergent for all a in [0,1[ , does it follow that the series \sum tn is also divergent"; thanks for your help $\endgroup$ – teller Aug 27 '18 at 10:01
  • $\begingroup$ BTW I'd guess that (ca.classical-analysis-and-odes) would be a more suitable top-level tag than (nt.number-theory). But probably it's better if I leave this up to you - I do not want to go overboard with editing your question. $\endgroup$ – Martin Sleziak Aug 27 '18 at 11:39
  • 3
    $\begingroup$ This isn't a Dirichlet series. $\endgroup$ – Greg Martin Aug 27 '18 at 18:59
5
$\begingroup$

What about $t_n=\frac{1}{n(\log n)^2}$?

$\endgroup$
  • 1
    $\begingroup$ thanks a lot; i tried 2 days to prove it (without success), now i know more:its false $\endgroup$ – teller Aug 27 '18 at 11:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.