Example 4.16. Investigate the convergence of the series $\sum_{n=1}^{\infty} 3^{(-1)^{n}-n}$
Solution. We apply the strengthened form of the D'Alembert's criterion (Theorem 4.10).

$$
\begin{aligned}
& \varlimsup_{n \rightarrow \infty} \frac{3^{(-1)^{n+1}-n-1}}{3^{(-1)^{n}-n}}=\varlimsup_{n \rightarrow \infty} 3^{(-1)^{n+1}-(-1)^{n}-1}=3 \\
& \varliminf_{n \rightarrow \infty} \frac{3^{(-1)^{n+1}-n-1}}{3^{(-1)^{n}-n}}=\underline{l i m}_{n \rightarrow \infty} 3^{(-1)^{n+1}-(-1)^{n}-1}=\frac{1}{27}
\end{aligned}
$$

These results do not allow us to draw any definite conclusions, i.e., even the strengthened form of D'Alembert's criterion does not provide a result. We apply the Cauchy radical criterion for positive series (Theorem 4.11)

$$
\lim _{n \rightarrow \infty} \sqrt[n]{3^{(-1)^{n}-n}}=\lim _{n \rightarrow \infty} \frac{\sqrt[n]{3^{(-1)^{n}}}}{3}=\frac{1}{3} \lim _{n \rightarrow \infty} 3^{\frac{(-1)^{n}}{n}}=\frac{1}{3}<1
$$

Therefore, the series under investigation converges. The examples 4.15 and 4.16 show that the Cauchy criterion is stronger than the D'Alembert's criterion.