Does the convergence of $|a_n|$ imply the convergence of $\sqrt[4]{\frac{|a_n|}{n^4}}$ ?
E.g. in R: $$\sum_{n=0}^{\infty} |a_n|\rightarrow a \in R \Rightarrow \sum_{n=0}^{\infty} \sqrt[4]{|\frac{a_n}{n^4}|}\rightarrow b \in R$$
Looks like the statement is true! According to comments we only needed to glance at something called Hölder's inequality. It states that:
$$\forall p \in (1,\infty) \,\exists\, q=\frac{p}{p-1} \,:\, \sum_{n=1}^{\infty}|b_n c_n|\leq \left(\sum_{n=1}^{\infty} |b_n|^p\right)^{1/p} \cdot \left(\sum_{n=1}^{\infty} |c_n|^q\right)^{1/q}$$
Thus, let $b_n:=\frac{1}{n}$ and $c_n:=|a_n|^{.25}$. We'll want to choose $p$ and $q$ such that the expression simplifies towards something more friendly and we can happily apply the comparison test. For this, it looks like we will want both terms to converge. We do not need to care about the exponents $1/p$ and $1/q$ of our new terms, since they'll be evaluated after taking the limit.
For the harmonic series to converge we need to marginally increase its exponent, that is, any $p > 1$ will do. Due to the $\frac{1}{p} + \frac{1}{q} = 1$ relation, we get to choose any $q \in (1,\infty)$. Yay! Lets try 4 in order to get rid of the quartic root. With $q=4$ and $p=\frac{4}{3}$ we get back |a_n|, which is by definition already guaranteed to converge :)
Apply Hölder's inequality with $x_k= \vert a_k\vert^{1/4},\ y_k= \frac{1}{k},\quad \frac{1}{p} = \frac{1}{4},$ and $\frac{1}{q} = \frac{3}{4}.$ This gets us,
$$ 0\leq \sum_{k=1}^{n} \sqrt[\Large{4}]{\frac{\vert a_k\vert}{k^4}} = \sum_{k=1}^{n} \vert a_k\vert^{1/4} \left(\frac{1}{k}\right) \leq \left(\sum_{k=1}^{n} \left( \vert a_k\vert^{1/4} \right)^4 \right)^{1/4} \left(\sum_{k=1}^{n} \left( \frac{1}{k} \right)^{4/3} \right)^{3/4}. $$
Both series in the final expression converge as $n\to\infty,$ so the result follows by the Sandwich theorem for sequences.