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.
 A: We have:
\begin{align}
\sqrt[4]{n^4+2n^2}
&=n\sqrt[4]{1+\frac 2{n^2}}\\
&=n\left(1+\frac 1{2n^2}+O\left(\frac 1{n^4}\right)\right)\\
&=n+\frac 1{2n}+O\left(\frac 1{n^3}\right)
\end{align}
Clearly, $p(n)\sim n^3$, hence $p(n)=n^3+an^2+bn+c$ and
\begin{align}
\sqrt[3]{p(n)}
&=n\sqrt[3]{1+\frac an+\frac b{n^2}+\frac c{n^3}}\\
&=n\left(1+\frac a{3n}+\frac b{3n^2}-\frac a{9n^2}+O\left(\frac 1{n^3}\right)\right)\\
&=n+\frac a3+\left(\frac b3-\frac a9\right)\frac 1n+O\left(\frac 1{n^2}\right)\\
\end{align}
Comparing we get $\frac a3=0$ and $\frac b3-\frac a9=\frac 12$ which gives $a=0$ and $b=\frac 32$, hence $p(n)=n^3+\frac 32n+c$ for all $c\in\Bbb R$.
A: $$\sqrt[4]{n^4+2n^2} = n\sqrt[4]{1+\frac{2}{n^2}}= n\left(1+\frac{1}{2n^2}-\frac{3+o(1)}{8n^4}\right) = n+\frac{1}{2n}-\frac{3+o(1)}{8n^3} $$
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 $\frac{6}{5}$.
