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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?

Analysis
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2 Answers2

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$2^n$ 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 $0$.

auscrypt
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    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... – Analysis May 22 '19 at 19:25
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    It isn't just larger, the growth rate is higher too (see https://en.wikipedia.org/wiki/Exponential_growth#Other_growth_rates) – auscrypt May 22 '19 at 19:28
  • Hi again, how do you know that the sequence is now bounded above by $\frac 1 {\sqrt{n}}$? – Analysis May 24 '19 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 '19 at 17:49
  • Ok, I really like that way since I have already proved that $2^n>n^2$ for all $n\in \mathbb{N}_{\geq 4}$ – Analysis May 24 '19 at 18:13
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Note that $$ \lim_{n\to \infty} \frac {a_{n+1}}{a_n}= 1/2 $$ thus your sequence tends to zero.

Mohammad Riazi-Kermani
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  • What about: $\left(\frac{n^{\frac{3}{2}}}{2^n}\right)^{\frac{1}{n}}=0.5\cdot n^{\frac{3}{2n}}= ....$ – Analysis May 24 '19 at 14:12