How to prove that $\lim_{n\to \infty} (n^k/2^n) = 0$? I'm having a hard time trying to prove this statement.
$\lim_{n\to \infty} (n^k/2^n) = 0$
k is a positive number.
Please, help me.
Thanks in advance.

Let us prove that $\lim_{n \to \infty}\frac{n^k}{a^n}=0,$ for $a>1$. Let $m \in Z$ and $m>k$, then
$$ 0<\frac{n^k}{a^n}\le \frac{n^m}{a^n}=\left(\frac{n}{\sqrt[m]{a^n}}\right)^m=\left(\frac{n}{b^n} \right)^m $$
where $b=\sqrt[m]{a}>1$.
Now,
\begin{align*} 0<\frac{n}{b^n}&=\frac{n}{(1+(b-1))^n}=\frac{n}{\sum\limits_{k=0}^n {n \choose k}1^{n-k}(b-1)^k}\\
&=\frac{n}{1+n(b-1)+\frac{n(n-1)}{2}(b-1)^2+\cdots+(b-1)^n} \\
&<\frac{n}{\frac{n(n-1)}{2}(b-1)^2} = \frac{2}{(b-1)^2(n-1)} \rightarrow 0 ~ (n \to \infty)
\end{align*}
Since $\frac{n}{b^n} \rightarrow 0 \Rightarrow (\frac{n}{b^n})^m \rightarrow 0~(n \to \infty)$
From $0<\frac{n^k}{a^n}\le \frac{n^m}{a^n}=\left(\frac{n}{\sqrt[m]{a^n}}\right)^m=\left(\frac{n}{b^n} \right)^m$, we have that $$\frac{n^k}{a^n} \rightarrow 0~ (n \to \infty) $$
