Given that, how would you evaluate the sequence given n approaches infinity.
I thought it would converge to 0 but that's wrong. I tried dividing the top and bottom by n!, resulting in 3/n! on the top and 2^n/n! on the bottom.
It seems like an indeterminate form just looking at it. The top would approach infinity and so would the bottom. So using L'hopitals rule, would deriving the top give me 0? It looks like a constant for any n; and deriving the bottom would give me:
n(2^n-1). Which also results in a convergence of 0.