Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$ I need help on a homework assignment. How to show that $\lim_{n\to\infty} \left(\dfrac{n^5}{3^n}\right) = 0$? We've been trying some things but we can't seem to find the answer.

Let $a_n = \frac{n^5}{3^n}$. Then clearly $a_n >0$ for all $n \in \mathbb{N}$. The ratio of two consecutive terms of the sequence is:
$$
   \frac{a_{n+1}}{a_n} = \frac{(n+1)^5}{3^{n+1}} \frac{3^n}{n^5} = \frac{1}{3} \left(1+\frac{1}{n}\right)^5
$$
For $n \geqslant 5$:
$$
  \frac{a_{n+1}}{a_n} = \frac13\left(1+\frac{1}{n}\right)^5 \leqslant \frac{1}{3} \left(1+\frac{1}{5}\right)^5 < 1
$$
Hence for $n\geqslant 5$ the sequence is decreasing, and since sequence elements are positive the sequence converges to this lower bound, i.e. zero.
