Answer
See explanation below.
Work Step by Step
Let ${a_k} = \dfrac{k}{{{3^k}}}$.
Condition 1. Examine if ${a_k} \gt {a_{k + 1}}$.
$\dfrac{k}{{{3^k}}} - \dfrac{{k + 1}}{{{3^{k + 1}}}} = \dfrac{k}{{{3^k}}} - \dfrac{k}{{{3^{k + 1}}}} - \dfrac{1}{{{3^{k + 1}}}} = \dfrac{1}{{{3^k}}}\left( {k - \dfrac{k}{3} - \dfrac{1}{3}} \right)$
$\dfrac{k}{{{3^k}}} - \dfrac{{k + 1}}{{{3^{k + 1}}}} = \dfrac{1}{{{3^k}}}\left( {\dfrac{2}{3}k - \dfrac{1}{3}} \right)$
For $k \ge 1$, we have $\dfrac{2}{3}k - \dfrac{1}{3} \gt 0$. Therefore,
$\dfrac{k}{{{3^k}}} - \dfrac{{k + 1}}{{{3^{k + 1}}}} \gt 0$
$\dfrac{k}{{{3^k}}} \gt \dfrac{{k + 1}}{{{3^{k + 1}}}}$
Hence, the condition ${a_k} \gt {a_{k + 1}}$ is satisfied.
Condition 2. Examine if $\mathop {\lim }\limits_{k \to \infty } {a_k} = 0$.
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{k}{{{3^k}}}$
Applying L'Hôpital's Rule, we get
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{k}{{{3^k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{\left( {\ln 3} \right){3^k}}} = 0$
Since the two conditions above are satisfied, by Theorem 9.6.1, the alternating series converges.
