Limit proof using ratio test I have been trying to prove the limit:
$$\lim_{k\to\infty} \frac{k^n}{2^k} = 0$$
for each $n\in\mathbb{N}$.
I was able to do a similar proof, specifically
$$\lim_{k\to\infty} \frac{k}{2^k} = 0$$
by using the ratio test.
For this problem, I think that I should be doing the same thing, but I get stuck on the last part. 
By starting with the ratio test for sequences:
For a sequence $(a_j)$ with the properties $r<1, K\in\mathbb{N}$, and $j\geq K \implies \left\lvert\frac{a_{j+1}}{a_j}\right\rvert\leq r$, it is true that $a_j\rightarrow0$ as $j\rightarrow 0$.
For my proof, I start with:
$$\left\lvert\frac{a_{k+1}}{a_k}\right\rvert = \frac{(k+1)^n}{k^n}\frac{2^k}{2^{k+1}}=\frac{1}{2}\left(\frac{k+1}{k}\right)^n$$
Now, if I could show that $\left(\frac{k+1}{k}\right)^n<2$, then I can complete the proof by showing that the ratio is less than 1 for all k greater than a certain value. It looks to me like $k\geq n+1$, but I don't know how to show this last part! In my analysis course, we have not yet defined the real numbers, otherwise I would be able to use the $n^{th}$ root of 2 to start.
I would appreciate any ideas! If there an easier way to prove this than the ratio test, it would be good to know about that as well.
 A: Alternatively, you may write, as $k \to +\infty$,
$$
\ln \left(\frac{k^n}{2^k} \right)=\ln \left(k^n\right)-\ln \left(2^k\right)=n\times \ln k-k \times \ln 2=k\left(n\:\frac{\ln k}{k} -\ln 2\right)\to +\infty\times (-\ln 2)
$$ thus
$$
\ln \left(\frac{k^n}{2^k} \right) \to -\infty
$$ and
$$
\frac{k^n}{2^k} \to 0.
$$
A: You can indeed prove that the sequence converges to $0$ via the ratio test, because in this case the series
$$
\sum_{k\ge0}\frac{k^n}{2^k}
$$
converges and so its general term has limit $0$. Your strategy is good:
$$
\frac{(k+1)^n}{2^{k+1}}\bigg/\frac{k^n}{2^k}=
\frac{1}{2}\left(1+\frac{1}{k}\right)^{\!n}
$$
Since $x\mapsto x^n$ is continuous and
$$
\lim_{k\to\infty}\left(1+\frac{1}{k}\right)=1
$$
we also have
$$
\lim_{k\to\infty}\frac{1}{2}\left(1+\frac{1}{k}\right)^{\!n}=
\frac{1}{2}<1
$$
A different strategy is to write $s=\sqrt[n]{2}$ so you have to compute
$$
\lim_{k\to\infty}\left(\frac{k}{s^k}\right)^n
$$
and the same argument as before shows you just need to show that
$$
\lim_{k\to\infty}\frac{k}{s^k}=0
$$
If you consider the limit of the function $f(x)=x/s^x$ you can apply l'Hôpital because
$$
\lim_{x\to\infty}s^x=\lim_{x\to\infty}e^{x\log s}=\infty
$$
when $s>1$; then you have
$$
\lim_{x\to\infty}\frac{x}{s^x}=
\lim_{x\to\infty}\frac{1}{s^x\log s}=0
$$
A: It's not a series so you can't really use the ratio test however if treat it like one and prove the series converges then that would mean the limit of the function is 0. 
Theorem 
If s_n converges then lim n->inf a_n =0
Now to prove this we just say 
Lim n->inf (k+1)^n/2^(k+1) * 2^k/k^n which simplifies to 
1/2 lim n->inf (k+1)^n/k^n which equals 1/2
Since this is less than 1 we can say the series 
S_n = k^n/2^k converges and thus the limit equeals zero
