# 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?

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

$$(n+1)^a=\sum_{j=0}^a\binom ajn^j<n^a+2^an^{a-1}$$

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 $b_n\to 0$.

• Thanks for this. Does this technique generalizes; i.e. if I have a series of the form $\frac{(a_n)_{n \in \mathbb{N}}}{(b_n)_{n \in \mathbb{N}}}$, applying the ratio test to the corresponding series will generally yield an answer? – Nagase Jan 29 '17 at 15:33

Root test should be enough to prove convergence for the series $\sum\limits_{n=0}^\infty \frac{n^a}{2^n},$ hence $\frac{n^a}{2^n}\!\rightarrow 0$.

In fact, $\limsup\limits_{n\to\infty}\sqrt[n]{|a_n|} = \lim\limits_{n\to\infty}\frac{n^{a/n}}{2} = \frac{1^a}2 = 1/2 < 1$, therefore the series converges, and the sequence must yield to zero.

• The second equality is given because $\lim\limits_{n\to\infty} n^{a/n} = 1^a$, right? – Nagase Jan 29 '17 at 15:35
• Yes exactly, since $\lim\limits_{n\to\infty} n^{1/n} = 1$ – Ottavio Bartenor Jan 29 '17 at 15:46