Determine if $\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$ converges or diverges. Another series I found I'm struggling with.

Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$

Ratio test and n-th root test are both inconclusive, Leibniz - criterion cannot be applied since the sequence given is not in the form of $(-1)^na_n$. I am sure the problem can be solved with the limit comparison test, though, the $n^{(...)}$ look pretty inviting after all. Let $a_n:=\frac{(-1)^nn^2+n}{n^3+1}$ then $|a_n|= \frac{n^2+n}{n^3+1}$. A try showing that the series diverges using the divergence of $\sum\frac{1}{n}$. For $n≥1$ it is clear that
$$ \frac{n^3+n^2}{n^3+1} ≥ 1 $$ dividing by $n$ yields:
$$ \frac{n^2+n}{n^3+1} = |a_n| ≥ \frac{1}{n}$$
Thus the series diverges. (?) 
EDIT: Hints you gave me yield:
$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1} = \sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1}  +
\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ With the first series converging by Leibniz-theorem and the second by limit comparison test with $\frac{1}{n^2}$.
 A: The series
$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1}$$ converges by Leibniz criterion, in addition the series
$$\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ converges by the comparison test.
Hence
$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} + \sum_{n=1}^{\infty}\frac{n}{n^3+1} = \sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$ also converges
A: Let denote
$$\frac{(-1)^nn^2+n}{n^3+1}=\underbrace{\frac{(-1)^nn^2}{n^3+1}}_{=u_n}+\underbrace{\frac{n}{n^3+1}}_{=v_n}$$
The series $\displaystyle \sum_n v_n$ is convergent since $v_n\sim_\infty \frac 1 {n^2}$.
We have 
$$u_n=\frac{(-1)^nn^2}{n^3+1}=\frac{(-1)^n}{n}\frac{1}{1+\frac{1}{n^3}}=\frac{(-1)^n}{n}\left(1-\frac{1}{n^3}+o\left(\frac{1}{n^3}\right)\right)$$ 
so we can see that $\displaystyle\sum_n u_n$ is a sum of convergent series hence it's convergent. Conclude.
A: Here is a different approach.
The sum of each pair of terms is
$$
\begin{align}
a_{2n-1}+a_{2n}
&=\frac{-(2n-1)^2+(2n-1)}{(2n-1)^3+1}+\frac{(2n)^2+2n}{(2n)^3+1}\\
&=\frac{4n^3-6n^2+5n-1}{n(4n^2-2n+1)(4n^2-6n+3)}\\
&=\frac1{4n^2}\frac{1-\frac3{2n}+\frac5{4n^2}-\frac1{4n^3}}{\left(1-\frac1{2n}+\frac1{4n^2}\right)\left(1-\frac3{2n}+\frac3{4n^2}\right)}\\
&=\frac1{4n^2}+O\left(\frac1{n^3}\right)
\end{align}
$$
Thus, the sum of $a_{2n-1}+a_{2n}$ converges by comparison to $\frac1{n^2}$.
Since the terms go to $0$, convergence of the partial sums is the same as convergence of every other partial sum.
Let $s_n=\sum\limits_{k=1}^n a_k$ and $p_n=\sum\limits_{k=1}^n(a_{2k-1}+a_{2k})$. Then $s_{2n}=p_n$ and $s_{2n+1}=p_n+a_{2n+1}$
Since $\lim\limits_{n\to\infty}a_n=0$, if $p_n$ converges, then $s_n$ converges.
Since $p_n$ is a subsequence of $s_n$, if $s_n$ converges, then $p_n$ converges.
That is, if $\lim\limits_{n\to\infty}a_n=0$, then
$$
\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty(a_{2n-1}+a_{2n})
$$
A: While the split-terms approach is easier, it's also worth noting that a little bit of massaging will let you apply the Leibniz criterion directly.  Consider $a_n=\dfrac{n^2+(-1)^nn}{n^3+1}$; then your series is of the form $\sum_{n=1}^\infty (-1)^na_n$.  Now, you should be able to prove that $a_n\gt 0$ for all $n$, and that $a_{n+1}\lt a_n$ for all sufficiently large $n$: for the latter, (fairly) obviously $a_{2k+1}\lt a_{2k}$, so you just have to show that $a_{2k}\lt a_{2k-1}$ for sufficiently large $k$.  Write out the terms explicitly, cross-multiply (since the denominators are manifestly positive) and cancel; you should find criteria on $k$ that force the inequality.
