# 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