Find polynomial $p(n)$ such that $\displaystyle \sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$ converges 
Find a polynomial $p(n)$ such that series:
$$\sum_{n=1}^{\infty} \big(\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\big)$$
converges.

Attempt. If $p(n)=q^3(n)$ for some polynomial $q(n)$ then
$$\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}= 
\sqrt[4]{n^4+2n^2}-q(n)=\frac{n^4+2n^2-q^4(n)}{(\sqrt[4]{n^4+2n^2}+q(n))\,(\sqrt{n^4+2n^2}+q^2(n))}$$
and I thought of getting $q(n)=n$, but in that case
$$\sqrt[4]{n^4+2n^2}-\sqrt[3]{p(n)}\sim \frac{1}{n}$$
and in that case we get divergence, by limit comparison test.
Thanks in advance.

$$\sqrt[4]{n^4+2n^2} = n\sqrt[4]{1+\frac{2}{n^2}}= n\left(1+\frac{1}{2n^2}-\frac{3+o(1)}{8n^4}\right) = n+\frac{1}{2n}-\frac{3+o(1)}{8n^3} $$
and $\left(n+\frac{1}{2n}\right)^3 = n^3+\frac{3n}{2}+o(1)$, hence by taking $\color{red}{p(n)=n^3+\frac{3n}{2}}$ we are fine, since
$$\sqrt[3]{n^3+\frac{3n}{2}}=n\sqrt[3]{1+\frac{3}{2n^2}}=n\left(1+\frac{1}{2n^2}-\frac{1+o(1)}{4n^4}\right) = n+\frac{1}{2n}-\frac{1+o(1)}{4n^3}$$
and $\sum_{n\geq 1}\frac{1}{n^3}=\zeta(3)$ is finite, close to $\frac{6}{5}$.
