0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $b_n\to 0$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Nagase Jan 29 '17 at 15:33
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.