real analysis - Convergence of a sequence $frac{n^a}{2^n}$

This is based on exercise 3.2.4 of D. J. H. Garling's A Course in Mathematical Analysis. Let $a \in \mathbb{N}$ be a fixed number and consider the sequence $(\frac{n^a}{2^n})_{n \in \mathbb{N}}$ (the exercise in Garling uses $a=10^6$). Does this sequence converge?

Heuristically, it seems to me clear that it does. Although $a$ may be very large, the sequence in the denominator grows faster than the one in the numerator, so that at some point for $n > a$, we will have a decreasing sequence. However, I've been unable so far to formalize this intuition in a rigorous argument. Can someone give me a hint, here?

Answer

Note that, for $a,n\ge 2$,

$$(n+1)^a=\sum_{j=0}^a\binom ajn^j

Then, if $b_n=n^a/2^n$ we have
$$0<\frac{b_{n+1}}{b_n}=\frac12\frac{(n+1)^a}{n^a}<\frac12\left(1+\frac{2^a}n\right)\to \frac12$$

So by ratio test, the series
$$\sum_{n=1}^\infty b_n$$
converges, and hence $b_n\to 0$.