How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}

How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}

prove that $$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<\dfrac{5}{3}$$
my idea: use $$3^k-2^k>2^k(k\ge 2)$$ then $$\Longleftrightarrow
1+\sum_{k=2}^{n}\dfrac{k}{3^k-2^k}<1+\sum_{k=2}^{n}\dfrac{k}{2^k}<1+\sum_{k=2}^{\infty}\dfrac{k}{2^k}=1+\dfrac{3}{2}>\dfrac{5}{3}$$
use this methods,I can't prove it
other idea: $$3^k-2^k>2\cdot 2^k(k\ge 3)$$
But This same can't prove it
so This problem have other nice methods? Thank you