Example 1. Find the interval of convergence of the series

$$
\sum_{n=1}^{\infty} \frac{(n+1)^{5}}{2 n+1}(x+1)^{n}
$$
Solution.

1. In this case, $c_{n}=\frac{(n+1)^{5}}{2 n+1} \neq 0$ for all $n$. Therefore, we can use formulas (1) or (2) for the radius of convergence of the power series.

By the D'Alembert's formula

$$
R=\lim _{n \rightarrow \infty}\left|\frac{c_{n}}{c_{n+1}}\right|=\lim _{n \rightarrow \infty} \frac{(n+1)^{5}}{2 n+1} / \frac{(n+2)^{5}}{2 n+3}=1
$$

Therefore, the interval of convergence is determined by the inequalities $-1<x+1<1$ and is given by $(-2,0)$.

2. We will investigate the convergence of the series at the boundary points of the interval of convergence. At the point $x=-2$, the power series takes the form

$$
\sum_{n=1}^{\infty} \frac{(n+1)^{5}}{2 n+1}(-1)^{n}
$$

and at the point $x=0$, the power series takes the form

$$
\sum_{n=1}^{\infty} \frac{(n+1)^{5}}{2 n+1}(1)^{n}
$$

Both series diverge because they do not satisfy the necessary condition for the convergence of series:

$$
\lim _{n \rightarrow \infty} \frac{(n+1)^{5}}{2 n+1}( \pm 1)^{n} \neq 0
$$

Therefore, the series diverges at these points.

Answer. The interval of convergence of the power series is $(-2,0)$.