lim3^nn^3 -3^nn!

Using the limit comparison test:

lim (3^n/n^3 - 3^n/n!) = lim (3^n/n^3) - lim (3^n/n!) = 0 - 0 (both are infinity/infinity form)

Let a_n = 3^n/n! Then, applying the ratio test:

lim |a_{n+1}/a_n| = lim |3^(n+1)/(n+1)! * n!/3^n| = lim |3/(n+1)| = 0 < 1

Therefore, the series converges by the ratio test and so the limit comparison test can be applied. Since the limit of both terms is 0, we can conclude that the original series also converges.