<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] We can write
$$\frac{n^2+3n-7}{n^3+1}= \frac{n^2-n +1}{n^3+1} + \frac{4(n +1)}{n^3+1} - \frac{12}{n^3+1}= \\
= \frac{1}{n+1} + \frac{4}{n^2-n +1} - \frac{12}{n^3+1}.
$$
Now, $\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n+1}}$ converges conditionally, however, $4\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n^2-n +1}}$ and $12\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n^3+1}}$ are absolutely convergent.
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