Notice that, for any $a>b>0$ and $n=1,2,\cdots$, it holds that
\begin{align*} a^n-b^n&=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})\\ &\geq (a-b)(b^{n-1}+b^{n-2}b+\cdots+bb^{n-2}+b^{n-1})\\ &=n(a-b)b^{n-1}\\ &\geq(a-b)b^{n-1}. \end{align*} Thus $$\frac{1}{3^n-2^n}\leq\frac{1}{2^{n-1}}.$$ Since $\sum\limits_{n=1}^{\infty}\dfrac{1}{2^{n-1}}$ is convergent,by the comparison test,$\sum\limits_{n=1}^{\infty}\dfrac{1}{3^n-2^n}$ is also convergent. But where does it converge to? WA gives a numerical approximation: $$\sum_{n=1}^{\infty}\frac{1}{3^n-2^n}\approx 1.27498.$$ I wonder whether it can be a closed form or not.
