Show that $\lim\limits_{n \to \infty} {\frac{n^s}{(1+p)^n}} = 0$ without using Ratio test. 
Problem: Let $s>0$, $p>0$. Show that $\lim\limits_{n \rightarrow \infty} {\displaystyle\frac{n^s}{(1+p)^n}} = 0$.

I can prove this problem by using Ratio test. But is there another way to prove this without using Ratio test?
 A: First show it is in indeterminate form, then apply l'Hopital's rule
$$\lim_{n \to \infty} \frac{n^s}{(1+p)^n}=\lim_{n \to \infty} \frac{sn^{s-1}}{(1+p)^n\ln(1+p)}=\frac{s}{\ln(1+p)}\lim_{n \to \infty} \frac{n^{s-1}}{(1+p)^n}$$
Keep applying L'Hopital's rule $n-1$ times, you will end up with a constant times $\lim_{n \to \infty}\frac{1}{(1+p)^n}$.  Just note that the constant is some power of $\frac{s}{\ln(1+p)}$ which is always a constant, so the whole expression will be $0$ as $n \to \infty$
A: You want a "direct proof" of it. Since the denominator is way larger than the numerator, you might want to involve squeeze theorem and try to establish the ...fact that: $\dfrac{n^s}{(1+p)^n} < \dfrac{1}{n}\implies (1+p)^n > n^{s+1}\iff n > a\ln n$ whereas $a > 0$ is a constant in terms of $p,s$. But you can write $n = e^{\ln n}  > 1 + \ln n + \dfrac{(\ln n)^2}{2}\implies n > \dfrac{(\ln n)^2}{2} > a\ln n$ when $n > e^{2a}$. Thus the inequality is validated and squeeze theorem goes through...
A: Let $a_n=\frac{n^s}{(1+p)^n}$. Then $a_n^{1/n} \to \frac{1}{1+p}$. Since $p>0$ there is $q$ such that $\frac{1}{1+p}<q<1$. Hence there is $N$ with
$a_n^{1/n}<q$ for all $n>N$. Hence
$0<a_n<q^n$ for all $n>N$, which gives the assertion.
A: $s=1:$ For $n>2,$ apply the binomial theorem in the denominator to get
$$\frac{n}{(1+p)^n} = \frac{n}{1+np + n(n-1)p^2/2 + \cdots } < \frac{n}{n(n-1)p^2/2}\to 0.$$
Now for any $s>0,$ we have $(1+p)^{1/s} > 1,$ which implies $(1+p)^{1/s} = 1+q$ for some $q>0.$ Thus the $s=1$ result implies
$$\frac{n^s}{(1+p)^n} = \left (\frac{n}{[(1+p)^{1/s}]^n}\right)^s  = \left (\frac{n}{(1+q)^n}\right)^s \to 0^s = 0.$$
