If $\sum_{n=1}^{\infty} a_{n}^{3}$ converges does $\sum_{n=1}^{\infty} \frac{a_{n}}{n}$ converge?

Solution 1:

$3|a_n|/n \leq (|a_n|^3 + (2/n^{3/2})$ 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.

Solution 2:

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 $\Sigma 1/n^{3/2}$.

Solution 3:

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).