Proving that $\lim_{n\rightarrow \infty} \frac{n^k}{2^n}=0$ I need to prove that $$\lim_{n\rightarrow \infty} \frac{n^k}{2^n}=0$$ where $k\in \mathbb{N}$. All I can think of is to use something like L'Hopital's rule but I suppose there must be a another simpler way. I would much appreciate if someone could give me a hint. Thanks
 A: Hint 1:-
$\displaystyle\lim_{n\rightarrow \infty} \dfrac{n^k}{2^n}=\displaystyle\lim_{n\rightarrow \infty}\left( \dfrac{n}{2^{\frac{n}{k}}}\right)^k$
Hint 2:-
$U_n=\dfrac{n^k}{2^n} \implies \displaystyle\lim_{n\to \infty}\dfrac{U_{n+1}}{U_n}=\displaystyle\lim_{n\to \infty}\dfrac{1}{2}\left(1+\dfrac{1}{n}\right)^k=\dfrac{1}{2}<1$
$\therefore \displaystyle\lim_{n\to \infty}U_n=0$
A: Here's a somewhat roundabout (but relatively elementary) approach. Fix $k\in\Bbb N$ and let $a_n:=\frac{n^k}{2^n}.$ Observe that $$\frac{a_{n+1}}{a_n}=\frac12\left(1+\frac1n\right)^k.$$ Show that there is some $N\in\Bbb N$ (dependent on $k$) such that $\left(1+\frac1n\right)^k<2$ whenever $n\ge N.$ Since each $a_n>0,$ and since $n\mapsto\left(1+\frac1n\right)^k$ is a decreasing function, then there is some $0<c<1$ such that for $n\ge N$ we have $\left(1+\frac1n\right)^k<2c,$ whence $a_{n+1}<ca_n$ for $n\ge N.$ Inductively, $0<a_{N+m}<c^ma_N$ for all $m\in\Bbb N.$ Apply Squeeze Theorem.
Let me know if you have trouble justifying any of these steps.
A: An alternative approach, you say?  By Bernoulli's inequality, when $n\ge k+1$ we have
$$ 2^n = \bigl((1+1)^{n/(k+1)}\bigr)^{k+1} \ge \Bigl(1+\frac n{k+1}\Bigr)^{k+1}
\ge \Bigl(\frac n{k+1}\Bigr)^{k+1} $$
Rearranging,
$$ \frac{n^k}{2^n} \le \frac{(k+1)^{k+1}}{n} \to 0
$$
A: $$\lim\limits_{n\to\infty}\frac{n^k}{2^n}=\left(\sqrt[k]{\lim\limits_{n\to\infty}\frac{n^k}{2^n}}\right)^k=\left(\lim\limits_{n\to\infty}\sqrt[k]{\frac{n^k}{2^n}}\right)^k=\left(\lim\limits_{n\to\infty}\frac{\sqrt[k]{n^k}}{\sqrt[k]{2^n}}\right)^k=\left(\lim\limits_{n\to\infty}\frac{n}{2^{\frac{n}{k}}}\right)^k=0^k=0$$
A: An alternative approach, you say?
$$ \frac{n^k}{2^n} = \frac{k^k}{2^n} \left(\frac nk\right)^k
\le \frac{k^k}{2^n}\binom nk \to 0
$$
For the fact that $\frac1{2^n}\binom nk\to 0$ as $n\to\infty$, see this question.  For the fact that $\bigl(\frac nk\bigr)^k\le\binom nk$, see this question.
(A slightly simpler variant is to say $\bigl(\frac nk\bigr)^k\le\binom nk\le 2^n$, and so $\frac{n^k}{2^n}\le k^k$.  But this is true for all $k$, so we also have $\frac{n^{k+1}}{2^n}\le (k+1)^{k+1}$, which yields $\frac{n^k}{2^n}\le\frac{(k+1)^{k+1}}{n}\to 0$.)
A: We have
$$2^n = e^{n\ln2} = \sum_{l=0}^{\infty} \dfrac{(n\ln2)^l}{l!} \geq \dfrac{(n \ln2)^{\lfloor k\rfloor+1}}{(\lfloor k\rfloor+1)!}$$
Hence,
$$\dfrac{n^k}{2^n} \leq \dfrac{(\lfloor k \rfloor+1)!}{\left(\ln(2)^{\lfloor k \rfloor+1}\right)} \dfrac1{n^{\lfloor k \rfloor-k+1}} \to 0$$
A: An alternative approach, you say?
$$ \sum_{n=0}^\infty \frac{n^k}{2^n}
= \left.\sum_{n=0}^\infty n^k x^n\right|_{x=\frac12}
= \left.\left(x\frac{d}{dx}\right)^k\sum_{n=0}^\infty x^n\right|_{x=\frac12}
= \left.\left(x\frac{d}{dx}\right)^k\frac1{1-x}\right|_{x=\frac12}
$$
By the quotient rule,
$$ \left(x\frac{d}{dx}\right)^k \frac1{1-x}
\quad\text{has the form}\quad \frac{\text{polynomial}}{(1-x)^{2^k}}
$$
and so it is finite for $x=\frac12$.  That is, $\sum_{n=0}^\infty \frac{n^k}{2^n}$ converges, so $\lim_{n\to\infty} \frac{n^k}{2^n}=0$.
