Proving convergence of $n^p/2^n$ as $n\to\infty$

Question

I was presented a problem stating that $n^{10000}/2^n$ converges as $n\to\infty$. I already know that there's a proposition that says the following;

Let $(\epsilon_k)^\infty_{n=0}$ be a null sequence of positive numbers. Then a sequence $(a_n)^\infty_{n=0}$ converges to $l$ if and only if for each $k$ there exists $n_k$ such that $|a_n-l|<\epsilon_k$ for all $n\geq n_k$.

I get kind of the idea of what is being said here, but how do I apply this to sequences in general, and in particular the problem stated above?

PS: is it also true that for any $p>0$ the same holds for $n^p/2^n$?

Answer 1

Note that by ratio test

$$a_n=\frac{n^p}{2^n}\implies \frac{a_{n+1}}{a_n}=\frac{(n+1)^p}{2^{n+1}}\frac{2^n}{n^p}=\frac12\left(\frac{n+1}{n}\right)^p\to \frac12<1\implies a_n\to 0$$

By the definition we need to prove that

$$0<\left|\frac{n^p}{2^n}-0\right|<\epsilon\iff 0<\frac{n^p}{2^n}<\epsilon \iff -\infty<p \log n-n\log 2<\log \epsilon$$

then all reduce to show that $f(x)=k\log x-x\to -\infty$ for $x\to \infty$.

Answer 2

Another way is to use $\ln(x)/x \to 0$ as $x \to \infty$ so $\ln(x)-x \to -\infty$.

Then, if $p>0$ and $a>1$, $n^p/a^n =e^{p\ln(n)-n\ln(a)} $. Since $\ln(a)>0$, $p\ln(n)-n\ln(a) \to -\infty$ as $n \to \infty$ so $n^p/a^n \to 0$.