Proof limit $b^n n^{\alpha} \to 0$, when $n \to \infty$ and $\forall |b| < 1 $. $$\lim_{n \to \infty} b^n n ^{\alpha} = 0, \forall |b| < 1, \alpha > 0$$
I have proved it for cases when $\alpha \le 0$, but I am not sure where to go further. I have tried D'Alembert's test, but I believe that it does not work for $\alpha > 0$. 
Can you give me any direction where to go ? At least, some nominal ideas.

An approach not using any kind of convergence criterion:
We may assume that $0<b<1$ (since, $b=0$ is trivial and convergence does not depend on sign of $b$).
Then we may write
$$
b=\frac{1}{1+r}
$$
where $r>0$ is real. Thus it follows that
$$
b^n=\frac{1}{(1+r)^n}=\frac{1}{1+nr+\cdots+{n \choose k} r^k+\cdots+r^n}
$$
We may assume that $n>\alpha+2$ (or just for sufficiently large $n$), then choosing $k=\lfloor{\alpha}\rfloor+1=\beta$ we have
$$
b^n=\frac{1}{1+nr+\cdots+{n \choose k} r^k+\cdots+r^n}\le \frac{1}{{n \choose \beta} r^{\beta}} = \frac{1}{\frac{n(n-1)\cdots (n-\beta+1)}{\beta !} r^{\beta}}
$$
so that
$$
b^n n^{\alpha} \le n^{\alpha} \frac{1}{\frac{n(n-1)\cdots (n-\beta+1)}{\beta !} r^{\beta}} \to 0
$$
as $n\to\infty$
