I can't figure out how to get this sequence to converge to 0 I'm trying to prove this sequence converges to $0$ (or maybe I'm wrong..):
$$k\in\mathbb N,1<q,\lim_{n\rightarrow\infty}\frac{n^k}{q^n}$$
I'd be happy to get some help with this! thank you!
 A: Let $a_n=\frac{n^k}{q^n}$.
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty} \frac{(n+1)^k}{q^{n+1}}/\frac{n^k}{q^n}=\frac{1}{q}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^k=\frac{1}{q}\left(1+\lim_{n\to\infty}\frac{1}{n}\right)^k=\frac{1}{q}$$
Hence there will be a $K$ after which the ratio is less than $b<1$ whenever $n\ge K$, so after that point it decreases exponentially. This is the same idea as in the proof of the ratio test, which further shows that your sequence, when summed, is also convergent (this also implies that the terms of the series go to 0).
Details
$\exists K\in\mathbb{N}$ s.t. $\forall n\ge K,\quad\frac{a_{n+1}}{a_n}<b$ where $b\in(\frac{1}{q},1)$.
$\forall n>K,\quad 0\le a_n=\frac{a_n}{a_{n-1}}a_{n-1}\le ba_{n-1}\le \dots\le b^{n-K}a_K\to 0$ as $n\to\infty$.
By squeeze theorem, $a_n\to 0$.
A: Hint. You are right. You may write, as $n$ is great:
$$
\log \left(\frac{n^k}{q^n}\right)=k \log n - n\log q=-n \left(\log q - k\frac{\log n}{n} \right)\sim -n \log q \longrightarrow -\infty
$$ thus your initial sequence is tending to $0^+$:
$$
\frac{n^k}{q^n}=e^{\log \left(\frac{n^k}{q^n}\right)} \longrightarrow e^{-\infty}=0
$$
A: $\frac{a_{n+1}}{a_n}=(\frac{n+1}{n})^k*\frac{1}{q}$. 
We have $q>1$ and $\lim_{n\to\infty}(\frac{n+1}{n})^k=1$.
Using the epsilon-delta definition, we have $N$ such that $(\frac{n+1}{n})^k<q\ \forall n>N$
Without loss of generality, assume it is so for all $n\in\mathbb{N}$. Thus, the sequence is decreasing, and we also have that it is bounded below by $0$. Thus, it is converging.
Denote the limit by $x$.$\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}$
$x=\frac{1}{q}x$ and thus $x=0$
A: So after long discussions about this problem I got a solution which doesn't use any of the suggested methods (such as limit of exponential's, logarithmic's, etc)
so my calculus course will accept it.
We find an $\alpha$ such that: (1) $\alpha^n>n^k$ $\land$ (2) $ 1\lt\alpha\lt q$
(1)
from the dense order of $\mathbb R$ and because $q>1$ then $\exists 1 \lt\alpha\lt q$
(2)
$\alpha^n>n^k \iff \alpha> \sqrt[n]{n}^k$ though $\lim\limits_{n \to \infty} \sqrt[n]{n}^k = 1^k = 1 \implies \exists N_0 \forall n>N_0 \,\mid\sqrt[n]{n}^k-1\,\mid<\frac{q-1}{2}$
from here
we choose $1<\frac{q-1}{2}<\alpha<q$ and we get $\frac{n^k}{q^n}<\frac{\alpha^n}{q^n}=(\frac{\alpha}{q})^n$ for $n>N_0$ though we know that
$\lim\limits_{n \to \infty} (\frac{\alpha}{q})^n = 0 \implies \lim\limits_{n \to \infty} \frac{n^k}{q^n} = 0$ from the squeeze theorem.
Wow! this is the first time I wrote something this long  here, thanks for the help!
