Example 4. Investigate the convergence of the series $\sum_{n=1}^{\infty} \frac{2^{n}(x+1)^{n}}{\left(3^{n}+2\right) n}$
Solution. Let $y=x+1$ and investigate the power series $\sum_{n=1}^{\infty} \frac{2^{n} y^{n}}{\left(3^{n}+2\right) n}$

1) $R=\lim _{n \rightarrow \infty} \frac{\left|a_{n}\right|}{\left|a_{n+1}\right|}=\lim _{n \rightarrow \infty} \frac{2^{n}}{\left(3^{n}+2\right) n} \cdot \frac{\left(3^{n+1}+2\right)(n+1)}{2^{n+1}}=\frac{3}{2}$.
2) When $y=\frac{3}{2}$, we obtain the numerical series $\sum_{n=1}^{\infty} \frac{3^{n}}{\left(3^{n}+2\right) n}$, which diverges.
3) When $y=-\frac{3}{2}$, we obtain the numerical series $\sum_{n=1}^{\infty} \frac{(-1)^{n} \cdot 3^{n}}{\left(3^{n}+2\right) n}$, which converges conditionally.

Thus, the power series, i.e., the series centered at zero $\sum_{n=1}^{\infty} \frac{2^{n} \cdot y^{n}}{\left(3^{n}+2\right) \cdot n}$, converges absolutely in the interval $\left(-\frac{3}{2} ; \frac{3}{2}\right)$, converges conditionally at the point $y=-\frac{3}{2}$, and diverges when $y \geqslant \frac{3}{2}$ and $y<-\frac{3}{2}$. Since $x=y-1$,
we will make the corresponding conclusion.

Answer. The given series converges absolutely in the interval $\left(-\frac{5}{2} ; \frac{1}{2}\right)$, converges conditionally at $x=-\frac{5}{2}$, and diverges when $x<-\frac{5}{2}$ and $x \geqslant \frac{1}{2}$.