Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{1}{3^k-2^k}$ and $b_k=\dfrac{1}{3^k}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{1}{3^k-2^k}}{1/3^k}\\=\lim\limits_{k \to \infty}\dfrac{3^k}{3^k-2^k} \\=\lim\limits_{k \to \infty}\dfrac{1}{1-(2/3)^k}\\=1$
Therefore, the series converges by the limit comparison test.
