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

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.

• Is $(a_n)$ assumed positive? – user370967 Jan 4 at 17:12
• Yes a_n >0 .Because other wise by alternating test directly we can argue. – MathLover Jan 4 at 17:13
• 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. – Mike Earnest Jan 4 at 17:19

Use Holder's inequality to deduce that $$\sum_{n} \frac{a_n}{n}\leq \left(\sum_n a_n^{3/2}\right)^{2/3}\left(\sum_n \frac{1}{n^3}\right)^{1/3}$$
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 $$\frac{a_n}n\le a_n^{3/2}$$.
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.
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.