0
$\begingroup$

This question already has an answer here:

Show that $\left(\frac{n^{\frac{3}{2}}}{2^n}\right)_{n\geq 0}$ is a null sequence. A null sequence is a sequence tending to $0$.

We need to find a $N\in \mathbb{N}$ for every $\varepsilon >0$, such that $n\geq N:|a_n-0|<\varepsilon$.

Usually, I first try to simplify the argument, but that does not work, since we have $n$ as the exponent and as a base. Secondly, I'll try to achieve an inequality like $n>...$. Tis sadly doesn't work out aswell. This expression is way to hard to simplify.

Is there another method, an easier way to solve this problem?

$\endgroup$

marked as duplicate by José Carlos Santos real-analysis May 22 at 19:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $2^n>n^2$ for $n>4$, so ... $\endgroup$ – user10354138 May 22 at 19:17
  • $\begingroup$ @robjohn huh? $0<n^{3/2}/2^n<n^{3/2}/n^2=n^{-1/2}$ etc. $\endgroup$ – user10354138 May 24 at 2:08
  • $\begingroup$ @user10354138: Sorry; I misread. I thought you were claiming that $2^n\gt n^\alpha\implies\frac{n^\alpha}{2^n}\to0$. $\endgroup$ – robjohn May 24 at 7:30
0
$\begingroup$

$2^n$ is exponential which grows faster than any power sequence. So at some point we must have $2^n>n^2$ (specifically $n\geq 4$ works), and so the sequence is now bounded above by $\frac{1}{\sqrt n}$, which clearly tends to $0$.

$\endgroup$
  • 1
    $\begingroup$ But how does showing that $2^n>n^2$ imply that it is a null sequence? You also have $n+1\geqslant n$ for each $n$, but $\left(\frac n{n+1}\right)_{n\in\mathbb N}$ is not a null sequence... $\endgroup$ – ParabolicAlcoholic May 22 at 19:25
  • 1
    $\begingroup$ It isn't just larger, the growth rate is higher too (see en.wikipedia.org/wiki/Exponential_growth#Other_growth_rates) $\endgroup$ – auscrypt May 22 at 19:28
  • $\begingroup$ Hi again, how do you know that the sequence is now bounded above by $\frac 1 {\sqrt{n}}$? $\endgroup$ – ParabolicAlcoholic May 24 at 17:47
  • $\begingroup$ @ParabolicAlcoholic We have $\frac{n^\frac{3}{2}}{2^n} < \frac{n^\frac{3}{2}}{n^2} = \frac{1}{\sqrt n}$. $\endgroup$ – auscrypt May 24 at 17:49
  • $\begingroup$ Ok, I really like that way since I have already proved that $2^n>n^2$ for all $n\in \mathbb{N}_{\geq 4}$ $\endgroup$ – ParabolicAlcoholic May 24 at 18:13
0
$\begingroup$

Note that $$ \lim_{n\to \infty} \frac {a_{n+1}}{a_n}= 1/2 $$ thus your sequence tends to zero.

$\endgroup$
  • $\begingroup$ What about: $\left(\frac{n^{\frac{3}{2}}}{2^n}\right)^{\frac{1}{n}}=0.5\cdot n^{\frac{3}{2n}}= ....$ $\endgroup$ – ParabolicAlcoholic May 24 at 14:12

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