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Show that $$\prod\limits_{k=1}^n\left(1-\frac1{2k}\right)<\frac1{\sqrt{2n+1}}$$ for every $n\ge 1$, essentially without induction

My attempt:

Let say $n$ is odd integer.

$$\displaystyle\prod_{k=1}^n\left(1-\frac1{2k}\right)=\dfrac1{2^n}\displaystyle\prod_{k=1}^n\left(\dfrac{2k-1}{k}\right)=\dfrac1{2^n}\left(\dfrac{1.3.5.7.9...(2n-1)}{1.2.3.4...(n-1).n}\right)=U$$

Hence:

$$U=\dfrac1{2^n}\left(\dfrac{\overbrace{(n+2)(n+4)...(2n-1)}^{Y}}{\underbrace{1.2.4.6...(n-3)(n-1)}_T}\right)$$

Let analyse $T$ ;

$$T=1.2.4.6...(n-3)(n-1)=2^{\frac{n-1}{2}}\left(1.2.3...\left(\frac{n-3}2\right)\left(\frac{n-1}2\right)\right)=2^{\frac{n-1}{2}}\left(\frac{n-1}{2}\right)!$$

and;

$$U=\dfrac1{2^n.2^{\frac{n-1}{2}}}\dfrac{Y}{\left(\frac{n-1}{2}\right)!}<\dfrac{(2n-1)^{\frac{n-1}2}}{2^{\frac{3n-1}{2}}\left(\frac{n-1}{2}\right)!}<\dfrac1{\sqrt{8^n-1}}$$

I think, It's not enough. How should we approach this inequality?

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  • $\begingroup$ In your very first step, shouldn't it be $\dfrac1{2^n}\displaystyle\prod_{k=1}^n\left(2-\frac1{k}\right)$? Then again I don't know if that changes any working after that. $\endgroup$ – John Doe Apr 16 '17 at 11:44
  • $\begingroup$ thanx, but it's only wrong written, rest of it, is still the same. $\endgroup$ – user2312512851 Apr 16 '17 at 11:46
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    $\begingroup$ This leaves open the case $n$ even, and, more importantly, this leaves us in a not very promising state (why are we supposed to believe the last inequality about U, one wonders). Thus the feeling that you did not advance one iota in the proof, actually... // Why the desire to avoid induction here? The proof by induction is a one-liner... $\endgroup$ – Did Apr 16 '17 at 11:56
  • $\begingroup$ This post of mine has a similar exercise too: math.stackexchange.com/a/1899918/321264. $\endgroup$ – StubbornAtom Apr 16 '17 at 12:03
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You want an overkill? You will get one. For first, we may notice that by integration by parts $$ \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right) = \frac{2}{\pi}\int_{0}^{\pi/2}\cos(x)^{2n}\,dx.\tag{1}$$ Then we may notice that over the interval $\left(0,\frac{\pi}{2}\right)$ we have $\cos(x)\leq e^{-x^2/2}$: it is enough to integrate, then exponentiating, both sides of the convexity inequality $-\tan(x)\leq -x$. It follows that:

$$ \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\leq \frac{2}{\pi}\int_{0}^{+\infty}e^{-nx^2}\,dx=\frac{1}{\sqrt{\pi n}}\tag{2} $$ that is stronger than the wanted inequality.

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  • $\begingroup$ This is nice, and gives ${2n \choose n} \le \frac{2^{2n}}{\sqrt{\pi n}}$ as a neat consequence. There's a case that an induction is hidden in the "integration by parts" section, but I don't really have a problem with this. $\endgroup$ – πr8 Apr 16 '17 at 14:40
  • $\begingroup$ @πr8: I agree, but one might avoid induction by turning the RHS of $(1)$ into a value of the Beta function. $\endgroup$ – Jack D'Aurizio Apr 16 '17 at 14:43
  • $\begingroup$ Thanks for the reply - yes, this seems like a reasonable enough workaround. $\endgroup$ – πr8 Apr 16 '17 at 14:45
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$$\prod\limits_{k=1}^n\left(1-\frac1{2k}\right)<\frac1{\sqrt{2n+1}}$$

$$\iff \prod\limits_{k=1}^n\left(1-\frac1{2k}\right)^{-2}>2n+1$$

$$\iff \prod\limits_{k=1}^n\left(\frac{2k}{2k-1}\right)^2>2n+1$$

Note that $$\left(\frac{2k}{2k-1}\right)^2=\left(1+\frac{1}{2k-1}\right)^2>1+\frac{2}{2k-1}=\frac{2k+1}{2k-1},$$ thus $$\prod\limits_{k=1}^n\left(\frac{2k}{2k-1}\right)^2 > \prod\limits_{k=1}^n\left(\frac{2k+1}{2k-1}\right) =2n+1$$ as required.

(if collapsing the telescoping product in the final step counts as "essentially induction", then I'll accept that judgment.)

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  • $\begingroup$ Induction: To be proved $$\left(1-\frac1{2n}\right)\frac1{\sqrt{2n-1}}<\frac1{\sqrt{2n+1}}$$ Algebraically equivalent to $$\sqrt{(2n-1)(2n+1)}<2n$$ End of the proof. $\endgroup$ – Did Apr 16 '17 at 12:43
  • $\begingroup$ @Did i agree that the problem is equivalent to the stated induction; i don't have much more to add to your comment. $\endgroup$ – πr8 Apr 16 '17 at 13:29

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