<s>[INST] Stuck on showing $\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$ converges I am trying to show if this series converges or diverges and I know it converges since for very large values of n, 
$$\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$$ becomes $$\sum_{n=0}^\infty \frac{(-1)^n1}{n}$$
which is convergent from the alternating series test since lim 1/n = 0 and 1/n is a decreasing sequence. I am going back and forth using the limit comparison test and alternating series test and can't seem to come up with anything. So right now i have $$\frac{(-1)^n(n^2+3n-7)}{n^3+1} \lt \frac{(-1)^n(n^2+3n-7)}{n^3}$$
and whenever I try to show the right handed side to be convergent I come back to the same problem as with the left hand side. 
 [/INST] I'll take the opportunity for showing some useful techniques for the manipulation of similar series.

Let $\omega=\frac{1+i\sqrt{3}}{2}$ and $\overline{\omega}=\omega^{-1}=\frac{1-i\sqrt{3}}{2}$. By the residue theorem
$$ \frac{x^2+3x-7}{x^3+1} = \frac{-3}{x+1}+\frac{2+\frac{2i}{\sqrt{3}}}{x-\omega}+\frac{2-\frac{2i}{\sqrt{3}}}{x-\overline{\omega}} \tag{1}$$
where:
$$ \sum_{n\geq 0}\frac{-3(-1)^n}{n+1} = -3\log(2)\tag{2} $$
so the original series equals
$$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n^2+3n-7)}{n^3+1} &=& -3\log(2)-4-\sum_{n\geq 0}\frac{(-1)^n 4n}{n^2+n+1}\\&\stackrel{ILP}{=}&-3\log(2)-4-2\int_{0}^{+\infty}\frac{\sqrt{3}\cos\left(\frac{s\sqrt{3}}{2}\right)-\sin\left(\frac{s\sqrt{3}}{2}\right)}{\sqrt{3}\cosh\left(\frac{s}{2}\right)}\,ds\\&=&-3\log(2)-4-\frac{4}{\sqrt{3}}\int_{0}^{+\infty}\frac{\sqrt{3}\cos\left(s\sqrt{3}\right)-\sin\left(s\sqrt{3}\right)}{\cosh\left(s\right)}\tag{3}\end{eqnarray*}$$
where ILP stands for Inverse Laplace Transform. By the Cauchy-Schwarz inequality, the absolute value of the last integral appearing in the RHS of $(3)$ is not larger than
$$ \frac{8}{\sqrt{3}}\int_{0}^{+\infty}\frac{ds}{\cosh s} = \frac{4 \pi}{\sqrt{3}}.\tag{4}$$
We may also simplify the RHS of $(3)$ a little more and state that:
$$ \sum_{n\geq 0}\frac{(-1)^n(n^2+3n-7)}{n^3+1} = -\log(8)-4-\frac{2\pi}{\cosh\left(\frac{\pi\sqrt{3}}{2}\right)}+\frac{4}{\sqrt{3}}\int_{0}^{+\infty}\frac{\sin(s\sqrt{3})}{\cosh(s)}\,ds.\tag{5}$$
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