Example 10. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{n^{2}+7}{n^{3}+3 n-2}$
Solution. Note that the necessary condition for convergence is satisfied:

$$
\lim _{n \rightarrow \infty} \frac{n^{2}+7}{n^{3}+3 n-2}=\lim _{n \rightarrow \infty} \frac{1+\frac{7}{n^{2}}}{1+\frac{3}{n^{2}}-\frac{2}{n^{3}}}=0
$$

However, this does not ensure the convergence of the series.

D'Alembert's ratio test:

$$
q=\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=\lim _{n \rightarrow \infty} \frac{(n+1)^{2}+7}{(n+1)^{3}+3(n+1)-2} \cdot \frac{n^{3}+3 n-2}{n^{2}+7}=1
$$

This result is consistent with both convergence and divergence of the series. The root test of Cauchy leads to the same conclusion (we suggest verifying this). Additional investigation is required.

It is not difficult to show that the improper integral $\int_{1}^{+\infty} \frac{x^{2}+7}{x^{3}+3 x-2} d x$ diverges, and thus the original series diverges (Theorem 9).

We can also apply the comparison test (Theorem 6). For this, we first determine the order of smallness of the general term of the series. We have:

$$
a_{n}=\frac{n^{2}+7}{n^{3}+3 n-2}=\frac{n^{2}\left(1+\frac{7}{n^{2}}\right)}{n^{3}\left(1+\frac{3}{n^{2}}-\frac{1}{n^{3}}\right)} \approx \frac{1}{n} \quad \text { as } \quad n \rightarrow+\infty
$$

We take the divergent harmonic series $\sum_{n=1}^{\infty} \frac{1}{n}$ as the series $\sum_{n=1}^{\infty} b_{n}$. From the obtained relation, it follows that $\lim \frac{a_{n}}{b_{n}}=1$. According to Theorem 6, the given series and the harmonic series behave the same in terms of convergence/divergence.

Answer. The series diverges.