Given a sequence $a_n$ such that $\sum_{n\ge1} \dfrac{|a_n|}{n^s}$ is convergent for $s>0$. Given that $\sum_{n\ge1} \frac{|a_n|}{n} < 1$, is it be possible to impose some sort of an upperbound for $\sum_{n\ge1} \frac{|a_n|}{n^s}$ for $1>s>k>0$ for some fixed $k$?

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    $\begingroup$ Given an integer $m$, define a sequence $a_n=\cases{m{\rm\ if\ }n=m\cr 0{\rm \ else}}$. While $\sum_{n\ge 1}{a_n\over n}=1$, you could replace $m$ with $m-\epsilon$ if you really cared about strict inequality. Since $\sum_{n\ge 1} {a_n\over n^s}=m^{1-s}$ and $m$ was chosen arbitrarily, there is no upper bound. $\endgroup$
    – B R
    Jan 26, 2011 at 2:45
  • $\begingroup$ [I deleted my earlier comment as it was based on a misreading of the desired range for $s$. BR's example/objection nails it in my opinion.] $\endgroup$
    – Yemon Choi
    Jan 26, 2011 at 3:08
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    $\begingroup$ I added my comment as an answer to keep the question from being bumped in the future. $\endgroup$
    – B R
    Jan 26, 2011 at 6:05
  • $\begingroup$ @BR - Thanks! Yeah, I wasn't really sure how to "accept" your answer, since you answered my question in a comment. $\endgroup$
    – BrainDead
    Jan 26, 2011 at 14:19

1 Answer 1


Given an integer $m$, define a sequence $a_n$ to be $m$ at the $m$-th place and $0$ otherwise. Then $$\sum_{n\ge 1} {a_n\over n^s}=m^{1-s}$$ As any upper bound must work for all $m$, we see that no upper bound is possible for $s<1$.


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