0
$\begingroup$

I have a small doubt and will much appreciate it if someone will clear it up. For large $k$ and $n=\lfloor 2^{k/2}\rfloor $, why are the following inequalities true:

  1. $\displaystyle\left(\frac{n}{2^{k/2}}\right)^k\le 1$.

  2. $\displaystyle\frac{2^{1+k/2}}{k!}$ is much smaller then 1.

Thanks a lot.

  • 1
    much smaller .... what exactly does that really mean?2012-05-26

2 Answers 2

1

Assuming that you made no typos, we have that $\lfloor 2^{k/2} \rfloor \leq 2^{k/2}$. So clearly

$ \dfrac{n}{2^{k/2}} \leq 1 $

and inequality (1) follows.

For inequality (2) just think about it in the following way:

$ \dfrac{2^k}{k!} = \dfrac{2\cdots 2}{1\cdots k} = \dfrac{2}{1} \cdot \dfrac{2}{2} \cdots \dfrac{2}{k} $ so when $k$ is large clearly the denominator is greater than the numerator. Since $1 + k/2 < k$ for $k$ sufficiently large, inequality $2$ follows.

1

For (2), suppose first that $k=2m$ is even. Then $\frac{k!}{2^m}=\frac{\big(1\cdot3\cdot5\cdot\ldots\cdot(2m-1)\big)(2\cdot4\cdot6\cdot\ldots\cdot 2m)}{2^m}=\big(1\cdot3\cdot5\cdot\ldots\cdot(2m-1)\big)m!\;,$ so $\frac{2^{1+k/2}}{k!}=\frac2{\big(1\cdot3\cdot5\cdot\ldots\cdot(2m-1)\big)m!}<\frac2{(k/2)!^2}\;,$ which is indeed much smaller than $1$ when $k$ is large. Only minor adjustments are required when $k$ is odd; I’ll leave them to you.

(1) is trivial as written: $0\le\lfloor 2^{k/2}\rfloor\le 2^{k/2}$ by the definition of floor and the fact that $2^{k/2}>0$, so $0\le\frac{\lfloor 2^{k/2}\rfloor}{2^{k/2}}\le 1\;,$ and therefore $0\le\left(\frac{\lfloor 2^{k/2}\rfloor}{2^{k/2}}\right)^k\le 1\;.$

If instead $n=2^{\lfloor k/2\rfloor}$, we have $0<2^{\lfloor k/2\rfloor}\le2^{k/2}$, and the same reasoning applies.