Show that $\left(\frac{n^{\frac{3}{2}}}{2^n}\right)_{n\geq 0}$ is a null sequence. [duplicate]

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?

marked as duplicate by José Carlos Santos real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 22 at 19:15

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

$$2^n$$ 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 $$0$$.
• 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... – ParabolicAlcoholic May 22 at 19:25
• Hi again, how do you know that the sequence is now bounded above by $\frac 1 {\sqrt{n}}$? – ParabolicAlcoholic May 24 at 17:47
• @ParabolicAlcoholic We have $\frac{n^\frac{3}{2}}{2^n} < \frac{n^\frac{3}{2}}{n^2} = \frac{1}{\sqrt n}$. – auscrypt May 24 at 17:49
• Ok, I really like that way since I have already proved that $2^n>n^2$ for all $n\in \mathbb{N}_{\geq 4}$ – ParabolicAlcoholic May 24 at 18:13
Note that $$\lim_{n\to \infty} \frac {a_{n+1}}{a_n}= 1/2$$ thus your sequence tends to zero.
• What about: $\left(\frac{n^{\frac{3}{2}}}{2^n}\right)^{\frac{1}{n}}=0.5\cdot n^{\frac{3}{2n}}= ....$ – ParabolicAlcoholic May 24 at 14:12