I would to find the smallest possible constant $c$ that satisfies

$$\frac{3^{3k}e\sqrt{3}}{\pi\sqrt{k}2^{3/2+2k}} \leq 2^{ck}$$

assuming $k\geq 1$ is an integer.

I tried taking logs base $2$ of both sides but that does not seem to be the right approach. How can you solve this problem?

When you do take log of both sides base $2$ you end up with this I think.

$$c \geq \frac{2+6k\ln{3} - 4k\ln{2} - 2\ln{\pi} + \ln{3} - 3\ln{2} - \ln{k}}{2k\ln{2}}$$

  • $\begingroup$ First, you'll want to factor out some constant terms of $$\frac{\sqrt 3 e}{\pi 2^{3/2}}$$ $\endgroup$ – AlexR Sep 21 '13 at 19:38
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    $\begingroup$ @Simon thats wrong; $\log$ is monotonously increasing, so all is fine. $\endgroup$ – AlexR Sep 21 '13 at 19:39
  • $\begingroup$ If we take a limit as $k\to \infty$, your final expression looks like $c\ge {6k(\ln 3 -\ln 2)+2k\ln 2\over 2k\ln 2}\to {3(\ln 3 -\ln 2)+\ln 2\over \ln 2}$. $\endgroup$ – abiessu Sep 21 '13 at 19:52
  • $\begingroup$ You need not to consider $\sup$. It is superfluous work. $\endgroup$ – user64494 Sep 21 '13 at 20:39

Taking natural logarithm, your inequality is equivalent to

$$ck\log2\geq3k\log3+1+\frac12\log3-\log\pi-\frac12\log k-\biggl(\frac32+2k\biggr)\log2\,,$$

and so

$$c\geq\frac1{\log2}\Biggl[3\log3-2\log2+\frac{1+\frac{\log3}2-\log\pi-\frac{3\log2}2}k-\frac{\log k}{2k}\Biggr]\,.$$


$$\begin{align*}c=&\,\sup_{k\in\mathbb Z^+}\frac1{\log2}\Biggl[3\log3-2\log2+\frac{1+\frac{\log3}2-\log\pi-\frac{3\log2}2}k-\frac{\log k}{2k}\Biggr]\\ =&\,\frac1{\log2}\Biggl[3\log3-2\log2+\frac12\,\sup_{k\in\mathbb Z^+}\biggl(\frac{a-\log k}k\biggr)\Biggr]\,, \end{align*}$$

where $a=2+\log3-2\log\pi-3\log2=\log\Bigl(\frac{3e^2}{8\pi^2}\Bigr)<0$.

If $f(x)=\frac{a-\log x}x$ for $x>0$, then $f'(x)=\frac{\log x-(a+1)}{x^2}$. Therefore

$$f'(x)>0\iff\log x>a+1\iff x>e\cdot e^a=\frac{3e^3}{8\pi^2}\,.$$

Since $\frac{3e^3}{8\pi^2}<1$, it follows that $f$ is increasing on the interval $[1,\infty)$. Therefore

$$\sup_{k\in\mathbb Z^+}\biggl(\frac{a-\log k}k\biggr)=\lim_{k\to\infty}\biggl(\frac{a-\log k}k\biggr)=0\,,$$

so your required value of $c$ is


  • $\begingroup$ You need not to consider $\sup$. It is a superfluous work. $\endgroup$ – user64494 Sep 21 '13 at 20:40
  • $\begingroup$ @user64494 I disagree. We are looking for the least value of $c$ satisfying the inequality for all the possible values of $k$. $\endgroup$ – Matemáticos Chibchas Sep 21 '13 at 21:16
  • $\begingroup$ @ Matematicos Schibchas: It is enough that a<0 and $$\lim_{k\to\infty}\biggl(\frac{a-\log k}k\biggr)=0. $$ $\endgroup$ – user64494 Sep 22 '13 at 10:01
  • $\begingroup$ @user64494 No, it is not enough, at least a priori: you also need to know that $(a-\log k)/k$ is increasing as function of $k$; otherwise the maximum value of this function could be located at an specific integer value. On the other hand, the functions $a/k, -1/k$ and $\log k$ are clearly increasing functions of $k$, so the same happens with our function $(a-\log k)/k$. $\endgroup$ – Matemáticos Chibchas Sep 22 '13 at 14:33

Making use of the Maple command $$sol := solve(3^{3*k}*exp(1)*sqrt(3)/(Pi*sqrt(k)*2^{3/2+2*k}) = 2^{c*k}, c) ,$$ I obtain $$sol=1/2\,{\frac {-4\,\ln \left( 2 \right) +6\,\ln \left( 3 \right) }{ \ln \left( 2 \right) }}+ $$ $$1/2\,{\frac {2+\ln \left( 3\,{\pi }^{-2} \right) +\ln \left( {k}^{-1 } \right) -3\,\ln \left( 2 \right) }{\ln \left( 2 \right) k}}. $$ In view of $$2+\ln \left( 3\,{\pi }^{-2} \right) -3\,\ln \left( 2 \right)=-1.270289025 $$ and $$1/2\,{\frac {-4\,\ln \left( 2 \right) +6\,\ln \left( 3 \right) }{ \ln \left( 2 \right) }}=2.754887504 $$ and taking into account that $\ln(\frac 1 k) \le0$ if $k \ge 1,$ I draw the conclusion that $c$ can be taken $2.754887504$.


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