Example 5.7. Using the D'Alembert's criterion, determine the convergence of the series

$$
\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\ldots+\frac{n}{3^{n}}+\ldots
$$
Solution. To use the D'Alembert's criterion, we need to know the ( $n+1$ )-th term of the series. It is obtained by substituting $n+1$ for $n$ in the expression for the general term of the series $a_{n}=\frac{n}{3^{n}}$: $a_{n+1}=\frac{n+1}{3^{n+1}}$. Now we find the limit of the ratio of the ( $n+1$ )-th term to the $n$-th term as $n \rightarrow \infty$:

$$
\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=\lim _{n \rightarrow \infty} \frac{n+1}{3^{n+1}}: \frac{n}{3^{n}}=\lim _{n \rightarrow \infty} \frac{(n+1) 3^{n}}{n \cdot 3^{n+1}}=\lim _{n \rightarrow \infty} \frac{n+1}{3 n}=\frac{1}{3} .
$$

Since $l=\frac{1}{3}<1$, the series converges.