2
$\begingroup$

I honestly have no idea how to solve it. I tried comparing the terms that are multiplied on each side but that didn't provide any use as far as I could tell. Also the $\frac{2n-1}{e}$ part seems so familiar to me - something to do with maximizing products given a constant sum I think? I'm not sure though and I can't find anything about it. If anyone can help me that'd be greatly appreciated

$\endgroup$
5
  • 1
    $\begingroup$ Have you tried induction? $\endgroup$
    – stange
    Nov 15, 2023 at 7:41
  • $\begingroup$ @stange I have but it didn't go anywhere. I'll try again and see what happens $\endgroup$
    – Marin
    Nov 15, 2023 at 7:44
  • $\begingroup$ possibly using Stirling inequalities for $\dfrac {(2n)!}{2^n\,n!}$ $\endgroup$ Nov 15, 2023 at 7:46
  • 1
    $\begingroup$ I would try to add up inequalities like $$2\ln(2k-1) < \int_{2k-1}^{2k+1} \ln(x) dx < 2\ln(2k+1).$$ $\endgroup$ Nov 15, 2023 at 8:15
  • 1
    $\begingroup$ Using @RaymondManzoni link and classical inequalities we need to show : $$\left(\frac{2x+1}{e}\right)^{\frac{2x+1}{2}}-\frac{\left(\frac{2x}{e}\right)^{2x}2\sqrt{\pi x}e^{\frac{1}{24x}}}{2^{x}\left(\frac{x}{e}\right)^{x}\sqrt{2\pi x}e^{\frac{1}{12x+1}}}>0,x\geq 1$$ math.stackexchange.com/questions/69162/… $\endgroup$ Nov 15, 2023 at 10:44

2 Answers 2

0
$\begingroup$

In $\frac{\left(2x\right)!}{2^{x}\left(x!\right)}$

If you write it as:

$A=$$\frac{\left(2x\right)!}{2^{x}\left(x!\right)}$$\left(\frac{e}{2x-1}\right)^{\frac{2x-1}{2}}$$\left(\frac{2x-1}{e}\right)^{\frac{2x-1}{2}}$ in which $M=\frac{\left(2x\right)!}{2^{x}\left(x!\right)}$$\left(\frac{e}{2x-1}\right)^{\frac{2x-1}{2}}$

$B=$$\frac{\left(2x\right)!}{2^{x}\left(x!\right)}$$\left(\frac{e}{2x+1}\right)^{\frac{2x+1}{2}}$$\left(\frac{2x+1}{e}\right)^{\frac{2x+1}{2}}$ in which $N=\frac{\left(2x\right)!}{2^{x}\left(x!\right)}$$\left(\frac{e}{2x+1}\right)^{\frac{2x+1}{2}}$

And check if $M > 1$

And check if $N < 1$, Then it can be proved.[Because We are comparing:$\left(\frac{2x-1}{e}\right)^{\frac{2x-1}{2}}$with M.$\left(\frac{2x-1}{e}\right)^{\frac{2x-1}{2}}$ AND $\left(\frac{2x+1}{e}\right)^{\frac{2x+1}{2}}$with N.$\left(\frac{2x+1}{e}\right)^{\frac{2x+1}{2}}$]

I don't know any algebraic way to do it, so I did it in 2 ways, first as we put [x = 1, M = 1.649, N = 0.863],[x = 2, M = 2.588, N = 0.654],[x = 3, M = 3.269, N = 0.547]... and so on. That means $M>1$ and $N<1$ for x = all positive integers. You can see that in Desmos also:

M(In Black) N(In Green)

enter image description here

$\endgroup$
1
  • $\begingroup$ This is not a proof just numerical verification for a range of values. $\endgroup$
    – Gary
    Nov 15, 2023 at 12:53
0
$\begingroup$

$$P_n=\prod_{i=1}^n(2i-1)=\frac{2^n }{\sqrt{\pi }}\,\Gamma \left(n+\frac{1}{2}\right)$$

For "large" values of $n$, using Stirling approximation $$\log(P_n)=n (\log (n)-1+\log (2))+\frac{\log (2)}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ $$\log(\text{lhs})=n (\log (n)-1+\log (2))-\frac{1}{2} (\log (n)+\log (2))+\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$ $$\log(\text{rhs})=n (\log (n)-1+\log (2))+\frac{1}{2} (\log (n)+\log(2))+\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$

Computing the difference of logarithms and exponentiating $$\frac{P_n}{\text{lhs}}=2 \sqrt{n}\left(1-\frac 1 {6n} +O\left(\frac{1}{n^2}\right)\right)$$ which is larger than one.

$$\frac{\text{rhs}}{P_n}=\sqrt{n}\left(1+\frac{1}{6 n} +O\left(\frac{1}{n^2}\right)\right)$$ which is larger than one.

I do not see how to use induction for this problem.

$\endgroup$
1
  • $\begingroup$ Yes I wasn't successful with induction so I don't think it's possible. $\endgroup$
    – Marin
    Nov 15, 2023 at 13:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .