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If $\sum a_n^{3/2}$ is convergent then $\sum \frac{a_n}n$ is convergent.

To prove this result: How to approach ?

Actually, when I first encountered this problem, I searched for an example that would counter this claim.

But I only have the usual example of harmonic series.

Hint only, please.

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    $\begingroup$ Is $(a_n)$ assumed positive? $\endgroup$ – user370967 Jan 4 at 17:12
  • $\begingroup$ Yes a_n >0 .Because other wise by alternating test directly we can argue. $\endgroup$ – MathLover Jan 4 at 17:13
  • $\begingroup$ Young's inequality $xy\le x^p/p+y^q/q$, valid for positive $x,y,p,q$ where $1/p+1/q=1$, is useful here. $\endgroup$ – Mike Earnest Jan 4 at 17:19
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Use Holder's inequality to deduce that $$ \sum_{n} \frac{a_n}{n}\leq \left(\sum_n a_n^{3/2}\right)^{2/3}\left(\sum_n \frac{1}{n^3}\right)^{1/3} $$

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One does not need to use Holder, Young, or anything else that advanced. The part of the series in the RHS corresponding to those $n$ with $a_n<\frac1{n^2}$ converges since $a_n<\frac1{n^2}$ yields $\frac{a_n}n<\frac1{n^3}$, the part corresponding to those $n$ with $a_n\ge\frac1{n^2}$ converges since in this case $\frac{a_n}n\le a_n^{3/2}$.

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Your thought of the harmonic series is a good one. If $a_n$ is nice enough, you can say $a_n^{3/2} \lt \frac 1n$, so $\frac {a_n}n \lt n^{-5/3}$. Of course, you were not given that $a_n$ is nice, so there is still work to do.

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  • $\begingroup$ This seems at most a comment. $\endgroup$ – Did Jan 4 at 17:36
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Note first that, for $a_1,\ldots,a_n\ge 0$, according to Hölder's inequality $$ \frac{a_1}{1}+\frac{a_2}{2}+\cdots+\frac{a_n}{n}\le (a_1^p+\cdots+a_n^p)^{1/p} \left(\frac{1}{1^q}+\cdots+\frac{1}{n^q}\right)^{1/q} $$ whenever $\displaystyle \frac{1}{p}+\frac{1}{q}=1$. In particular, for $p=3/2$ and $q=3$, we have $$ \frac{a_1}{1}+\frac{a_2}{2}+\cdots+\frac{a_n}{n}\le (a_1^{3/2}+\cdots+a_n^{3/2})^{2/3} \left(\frac{1}{1^3}+\cdots+\frac{1}{n^3}\right)^{1/3}\le \left(\sum_{n=1}^\infty a_n^{3/2}\right)^{2/3}\left(\sum_{n=1}^\infty\frac{1}{n^3}\right)^{1/3}=M<\infty. $$ Hence, $\sum_{n=1}^\infty\frac{a_n}{n}$ converges.

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