Prove that $\lim \limits_{n\to \infty}\frac{n^{\alpha}}{(1+p)^n}=0$ Prove that if $p>0$ and $\alpha\in \mathbb{R}$ then $$\lim \limits_{n\to \infty}\dfrac{n^{\alpha}}{(1+p)^n}=0.$$
No ideas how to prove it. Can anyone help please?
Prove without using logarithm.
 A: Consider $\sum_{i=1}^\infty \frac{n^\alpha}{(1+p)^n}$, and perform the Ratio Test for convergence of this series:
$$\lim_{n\to \infty} \left| \frac{(n+1)^\alpha}{(1+p)^{n+1}} \cdot\frac{(1+p)^n}{n^\alpha}\right| = \lim_{n\to \infty}\left|\left(1 + \frac{1}{n}\right)^\alpha \frac{1}{1+p} \right| = \frac{1}{1+p} < 1$$
As this limit is less than 1, the series is convergent, which is only possible if the sequence $\frac{n^\alpha}{(1+p)^n} \to 0$ as $n \to \infty$.
A: Let $u_n = \frac{n^\alpha}{(1+p)^n}$.
we have $u_{n+1}=u_n \times\frac{(n+1)^\alpha}{n^\alpha (1+p)}$.
It is sufficient that $\frac{(n+1)^\alpha}{n^\alpha (1+p)} < 1$ and this occurs
if $(n+1)^\alpha < n^\alpha (1+p)$ which implies $n+1 < n (1+p)^{1/\alpha}$.
This last inequality is true when n is sufficiently great ($n> n_0$) as $(1+p)^{1/\alpha}>1$.
So the sequence is decreasing to 0 for all $n > n_0$.
($n_0$ is function of $p$). 
A: Take the logarithm and show that it goes to -infinity. If the logarithm goes to -infinity, then the sequence itself goes to 0. 
