Prove that if $a,b\in \mathbb{R}$, $a>0$, then $\lim \frac{n^b}{(1+a)^n}=0$ as $n\to\infty$ As the title says, i want to show that $$\lim_{n\to \infty} \frac{n^b}{(1+a)^n}=0$$ where $a,b\in \mathbb{R}$ with $a&gt;0$.

I tried to bound the sequence $x_n =  \frac{n^b}{(1+a)^n}$ and use the sandwish theorem, but have no results. My problem is the exponent in the denominator, maybe it could be bounded with Bernoulli inequality ($(1+a)^n\geq 1+na$) but then the problem is de numerator $n^b$.

I'm grateful for any hints for doing this problem, or some steps to clarify.
HINT:

Bernoulli's inequality won't suffice when $b\ge1$, but the binomial theorem will suffice.

From the binomial theorem, with $n&gt;b$, we see that

$$(1+a)^n\ge \binom{n}{\lfloor b\rfloor +1}a^{\lfloor b \rfloor +1}$$

where $\lfloor \cdot \rfloor$ is the floor function.

Can you proceed?
By the ratio test, the sequence converges if
$$\frac{(n+1)^b}{(1+a)n^b}&lt;1.$$

This is achieved when 

$$n&gt;\frac1{\sqrt[b]{1+a}-1},$$ a finite number.



E.g., with $\dfrac{n^3}{2^n}$, for $n&gt;3$,

$$\frac{(n+1)^3}{2n^3}&lt;\frac{125}{2\cdot64}&lt;1.$$