Just checking. $2^n$ ($n \to \infty$) tends to $\infty$.
+ $3^n$ also ($n \to \infty$) tends to $\infty$
so the sum gets me $\infty$.
Now $(\infty)^{1/\infty}$ : $(\infty)^0 = 1$
I see no other way. Theorem: $n^{\frac{1}{n}} = 1$ as $n \to \infty$.
Analogous: $(2^n - 3^n)^\frac{1}{n}$ = $(\infty - \infty)^0$ = 1 ???
Is it ok to suppose infinity on any integer $k^n$ as $n$ goes to infinity ?