1000. Determine the domain of convergence of the functional series:

$$
\text { 1) } \sum_{n=1}^{+\infty} \frac{1}{n(x+2)^{n}} ; \quad \text { 2) } \sum_{n=1}^{+\infty} n \sqrt[3]{\sin ^{n} x}
$$
Solution: 1) Using the D'Alembert's criterion:

$$
\begin{aligned}
& u_{n}=\frac{1}{n(x+2)^{n}} ; u_{n+1}=\frac{1}{(n+1)(x+2)^{n+1}} ; \\
& \rho=\lim _{n \rightarrow+\infty}\left|\frac{u_{n+1}}{u_{n}}\right|=\lim _{n \rightarrow+\infty} \frac{n}{(n+1)|x+2|}=\frac{1}{|x+2|} . \\
& \frac{1}{|x+2|}<1 ; \quad |x+2|>1 ; \quad x+2>1 \text{ or } x+2<-1 ; \quad x>-1 \text{ or } x<-3 \\
& -\infty<x<-3, \quad -1<x<+\infty .
\end{aligned}
$$

The boundaries of the two found intervals are investigated separately.

For $x=-3$, we get an alternating numerical series with the general term $\frac{1}{n(-1)^{n}}$, which converges according to the Leibniz criterion.

For $x=-1$, we get a divergent harmonic series.

Therefore, the domain of convergence of the given series consists of two infinite intervals $(-\infty,-3]$ and $(-1,+\infty)$.
2) $u_{n}=n \sqrt[3]{\sin ^{n} x} ; \quad u_{n+1}=(n+1) \sqrt[3]{\sin ^{n+1} x}$;

$$
\rho=\lim _{n \rightarrow+\infty}\left|\frac{u_{n+1}}{u_{n}}\right|=\lim _{n \rightarrow+\infty} \frac{n+1}{n}|\sqrt[3]{\sin x}|=|\sqrt[3]{\sin x}|
$$

This limit $\rho$ will be less than one, and the series will be convergent according to D'Alembert's criterion, for all values of $x$ except $x_{k}=\frac{\pi}{2}+\pi k, k=0, \pm 1, \pm 2, \ldots$, for which $\rho=1$.

For $x=x_{k}$ and even $k$, we get the series $1+2+3+\ldots+n+\ldots$, and for odd $k$ the series $-1+2-3+\ldots+(-1)^{n} n+\ldots$, both of which diverge due to the non-satisfaction of the necessary condition for convergence.

Therefore, the domain of convergence of the given series is the entire number line, excluding the points $x_{k}$.