Find if the serie $\frac{n-1}{2n^3-n+4}$ converges. For doing this I chose to compare it to series $\frac{1}{n^2}$:$$\frac{n-1}{2n^3-n+4}\le\frac{n}{n^3}=\frac{1}{n^2}$$. And because $\Sigma\frac{1}{n^2} $ converges absolutely I can deduct that the serie $\Sigma\frac{n-1}{2n^3-n+4}$ also converges. I would like to ask whether the conclusion is true and also what would be the other good ways to determine in this instance whether the serie converges?
 A: Yes,your conlusion is true,
Another good general way,but in this case quite equivalent way to solve the exercises, is through ''asymptotic comparison'' with $b_{n} = \frac{1}{n^{2}}$.
Let's enounce it :
Let $\{a_{n}\},\{b_{n}\} \subset \mathbb{R}$ such that $\exists n_{0} \in \mathbb{N} : \forall n \geq n_{0} \{a_{n}\},\{b_{n}\} > 0$ and let's suppose exist $$\exists \lim\limits_{n\to +\infty}\frac{a_{n}}{b_{n}} = L \in [0,+\infty]$$
Then if $\begin{cases}L \in (0,+\infty) \rightarrow \sum\limits_{n \in \mathbb{n}}a_{n} \approx \sum\limits_{n \in \mathbb{n}}b_{n} \\ L=0 \sum\limits_{n \in \mathbb{n}} b_{n} < \infty \rightarrow \sum\limits_{n \in \mathbb{n}} a_{n} < \infty \\ L=+\infty \sum\limits_{n \in \mathbb{n}} b_{n} = +\infty \rightarrow \sum\limits_{n \in \mathbb{n}} a_{n} = +\infty\end{cases}$
In this case $a_{n}=\frac{n-1}{2n^{3}-n+4},b_{n} = \frac{1}{n^{2}}$ 
We can notice that $\lim\limits_{n\to +\infty}\frac{a_{n}}{b_{n}} = \frac{1}{2}=L \in (0,+\infty)$ so that $\sum\limits_{n \in \mathbb{n}} a_{n} < \infty$
A: Your approach is correct and in fact you have used the Comparison Test. Other useful tests can be found in https://en.wikipedia.org/wiki/Convergence_tests.
