If $\sum_{n=1}^{\infty} a_{n}^{3}$ converges does $\sum_{n=1}^{\infty} \frac{a_{n}}{n}$ converge? Suppose $a_{n}>0$ and the following series converges
$\sum_{n=1}^{\infty} a_{n}^{3}$
Does this imply that
$\sum_{n=1}^{\infty} \frac{a_{n}}{n}$ 
converges?
I was able to prove that the second series also converges by using the limit comparision test.   Is there another way to show the second series converges (e.g. root or ratio test)?
 A: Here's a more conceptual answer that doesn't resort to inequalities ad hoc.  Assume $a_n > 0$. 
The problem can be rephrased, by writing $(a_n)^3 = 1/(n^{3/2 + p})$ (with $p$ a function of $n > 1$, and ignoring $a_1$), as:
If $\Sigma 1/(n^{3/2 + p})$ converges then $\Sigma 1/(n^{3/2 + (p/3)})$ converges.  
The transformation moves the series toward the (convergent) one with $p=0$.  This preserves convergence, because the series can be split into the terms with $p \leq 0$ and $p>0$.  For the first set, convergence is improved, and for the second set, the sum is dominated by $\Sigma 1/n^{3/2}$. 
A: By Hölder's inequality you have
$$ A_k=\sum_ {i=1}^{k} \frac{a_n}{n} \leq \left ( \sum_ {i=1}^{k}a_ n ^3 \right )^\frac{1}{3} \left ( \sum_ {i=1}^{k} \frac{1}{n^\frac{3}{2}}\right ) ^\frac{2}{3}$$
By taking $k \rightarrow \infty $,we see $\sum_ {i=1}^{\infty} \frac{a_n}{n}$ is bounded and since $\frac{a_n}{n} \geq 0$ we conclude it is convergent( since $A_k$ is bounded and increasing).
A: $3|a_n|/n \leq (|a_n|^3 + (2/n^{3/2})$ by a well known inequality.  Sum.
This also shows the range of values that the second infinite sum can assume, given the value of the first, and assuming all terms are non-negative.
