Finding $\sup$ and $\inf$ of $\frac{n^5}{2^n}$ where $n$ is natural number I'm trying to find $\sup A, \inf A$ where
$$A=\{a_n=\frac{n^5}{2^n}:n\in\Bbb{N}\}, 1\not\in\Bbb{N}$$
For $n=1$ we have $a_1 = \frac{1}{2}$, $\lim_{x\rightarrow +\infty} \frac{n^5}{2^n}=0$ and after differentiating I found out that the critical point is at $n=\frac{5}{\ln 2}$. And there is my problem:
We know that the candidates for $\sup$ and $\inf$ are $0,\frac{1}{2}$ and $a_k, a_j$ for natural $k$, $j$ near $\frac{5}{\ln 2}$. But how to find $k$ and $j$? Clearly $\frac{5}{\ln 2}\not\in\Bbb{N}$ so $k=\lfloor\frac{5}{\ln 2}\rfloor$ and $j=\lceil\frac{5}{\ln 2}\rceil$ but how to find where $k$ and $j$ are exactly?

Note that the critical value $\frac{5}{\ln 2}\in(7,8)$ and hence $k=\lfloor\frac{5}{\ln 2}\rfloor=7$ and $j=\lceil\frac{5}{\ln 2}\rceil=8$.  But $a_7>a_8$ and so $a_1<a_2<\cdots<a_7>a_8>a_9>\cdots$ and hence $\sup\{a_n\}=a_7$ and $\inf\{a_n\}=0$.
