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 $k$?

3$\begingroup$ 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 $\sum_{n\ge 1} {a_n\over n^s}=m^{1s}$ and $m$ was chosen arbitrarily, there is no upper bound. $\endgroup$– B RJan 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 ChoiJan 26, 2011 at 3:08

1$\begingroup$ I added my comment as an answer to keep the question from being bumped in the future. $\endgroup$– B RJan 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. $\endgroup$– BrainDeadJan 26, 2011 at 14:19
1 Answer
Given an integer $m$, define a sequence $a_n$ to be $m$ at the $m$th place and $0$ otherwise. Then $$\sum_{n\ge 1} {a_n\over n^s}=m^{1s}$$ As any upper bound must work for all $m$, we see that no upper bound is possible for $s<1$.