# If $\sum_{n=1}^{\infty} a_{n}^{3}$ merges does $\sum_{n=1}^{\infty} \frac{a_{n}}{n}$ merge?

Intend $a_{n}>0$ and also the adhering to collection merges

$\sum_{n=1}^{\infty} a_{n}^{3}$

Does this indicate that

$\sum_{n=1}^{\infty} \frac{a_{n}}{n}$

merges?

I had the ability to confirm that the 2nd collection additionally merges by utilizing the restriction comparision examination. Exists an additional means to show the 2nd collection merges (e.g. origin or proportion examination)?

Here's an extra theoretical solution that does not consider inequalities impromptu. Think $a_n > 0$.

The trouble can be put in other words, by creating $(a_n)^3 = 1/(n^{3/2 + p})$ (with $p$ a function of $n > 1$, and also overlooking $a_1$), as :

*If $\Sigma 1/(n^{3/2 + p})$ merges after that $\Sigma 1/(n^{3/2 + (p/3)})$ converges *.

The makeover relocates the collection towards the (convergent) one with $p=0$. This maintains merging, due to the fact that the collection can be divided right into the terms with $p \leq 0$ and also $p>0$. For the first set, merging is boosted, and also for the 2nd set, the amount is controlled by $\Sigma 1/n^{3/2}$.