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Prove that $$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$

saw some similar problems like show $\frac{1}{15}< \frac{1}{2}\times\frac{3}{4}\times\cdots\times\frac{99}{100}<\frac{1}{10}$ is true

but didn't manage to get 115. I could get a weaker conclusion of $x>100$ though.

\begin{align} x^2 &= \left(\frac 21 \times \frac 21\right) \times \left(\frac 43 \times \frac 43\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{9999}\right) \\ &\ge \left(\frac 21 \times \frac 32\right) \times \left(\frac 43 \times \frac 54\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10001}{10000}\right) \\ &= 10001 \end{align}

so $x > 100$

Matt Frank
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    All people here are not for answering. Most are for learning. So even if you could prove that it is greater than 100, you should never hesitate to show up your work. This would be helpful to people like me also. – Nandeesh Bhatrai Feb 04 '21 at 20:47
  • Showing your work will also help people to write answers suitable to your level. – saulspatz Feb 04 '21 at 20:48
  • thanks guys, I put up my work of proving $x>100$ – Matt Frank Feb 04 '21 at 20:56
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    Your squaring method with a tweak on your first term could get there... you have a spare 4/3 lying around from approximating 2/1 by 3/2 – Joffan Feb 04 '21 at 21:01
  • You can also get better approximations by pulling out the first few terms and applying the squaring method to the rest. Writing $x=\frac21\times\frac42\times\frac65y$ and applying the squaring mathod to $y$, I got $x>\frac{320}{\sqrt7}\approx120.9486$ – saulspatz Feb 04 '21 at 21:19

5 Answers5

3

Looking at your method, you've actually proven $\frac{3}{4}x^2 > 10000$ as well – just, instead of writing $\frac{2}{1} \geqslant \frac{3}{2}$, you can incorporate that $\frac{3}{4}$ here, and you'll have exactly $\frac{3}{2} = \frac{2}{1} \cdot \frac{3}{4}$. Now we can deduce that

$$ x > \sqrt{\frac{40000}{3}} \approx 115.47 > 115$$

radekzak
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2

Show this product is equal to:

$$\frac{4^{5000}}{\binom{10000}{5000}}$$

Then use the inequality about the central binomial coefficient:

$$\frac{4^n}{\sqrt{4n}}\leq\binom{2n}{n}\leq\frac{4^n}{\sqrt{3n+1}}$$

Setting $n=5000$ this gives:

$$\frac{4^{5000}}{\sqrt{20000}}<\binom{10000}{5000}<\frac{4^{5000}}{\sqrt{15001}}$$

Or

$$122<\sqrt{15001}<\frac{4^{5000}}{\binom{10000}{5000}}<\sqrt{20000}<142$$

Thomas Andrews
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  • Hmm, I preferred radekzak’s answer. This one gets a better lower bound, but requires non-trivial results, while radekzak’s is just a tweak to your elementary result. – Thomas Andrews Feb 05 '21 at 21:17
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We have $$\dfrac{2}{1}\cdot \dfrac{4}{3}\cdots \dfrac{10000}{9999}= \dfrac{10000!!}{9999!!}= \dfrac{2^{5000}\cdot 5000!}{\dfrac{10000!}{2^{5000}\cdot 5000!}}= 2^{10000}\cdot \dfrac{\left ( 5000! \right )^{2}}{10000!}$$ Using Sterling's approximation $n!= \sqrt{2n\pi}\left ( \dfrac{n}{e} \right )^{n}$ then we have $$x\cong 2^{10000}\cdot \dfrac{\left ( \sqrt{10000\pi}\left ( \dfrac{5000}{e} \right )^{5000} \right )^{2}}{\sqrt{20000\pi}\left ( \dfrac{10000}{e} \right )^{10000}}= 2^{10000}\cdot \dfrac{50\sqrt{2\pi}}{2^{10000}}= 50\sqrt{2\pi}$$ Another solution is $$x= \frac{2^{10000}}{\dbinom{10000}{5000}}= \frac{2^{10000}}{\dfrac{2^{10001}}{\pi}\int_{0}^{\infty}\dfrac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}}= \frac{\pi}{2\int_{0}^{\infty}\dfrac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}}$$ Then by the transformation $\int_{0}^{\infty}f\left ( x \right ){\rm d}x= \int_{0}^{1}\dfrac{f\left ( \dfrac{x}{1- x} \right )}{\left ( 1- x \right )^{2}}{\rm d}x$ we have $$\int_{0}^{\infty}\frac{{\rm d}x}{\left ( x^{2}+ 1 \right )^{5001}}= \int_{0}^{1}\frac{{\rm d}x}{\left ( \left ( \dfrac{x}{1- x} \right )^{2}+ 1 \right )^{5001}\left ( 1- x \right )^{2}}= \int_{0}^{1}\frac{\left ( x- 1 \right )^{10000}}{\left ( 2x^{2}- 2x+ 1 \right )^{5001}}{\rm d}x=$$ $$= \int_{0}^{1}\frac{\left ( \dfrac{x^{2}- 2x+ 1}{2x^{2}- 2x+ 1} \right )^{5000}}{2x^{2}- 2x+ 1}{\rm d}x$$ We can doublecheck the result then in Wolfram|Alpha or Mathematica.

0

Working a little differently, you get an exact result: $$ \begin{align} x &= \left(\frac 21 \times \frac 22\right) \times \left(\frac 43 \times \frac 44\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{10000}\right) \\ &= \frac{1}{10000\, !} [2\times 4 \times \cdots \times 10000]^2\\ &= \frac{1}{10000\, !} [2^{5000}\times 1\times 2 \times \cdots \times 5000]^2\\ &= \frac{1}{10000\, !} 4^{5000}[5000 \, !]^2\\ &= \frac{4^{5000}}{\binom{10000}{5000}} \end{align} $$ Now proceed as in @Thomas Andrews's solution above.

Andreas
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You can make the problem more general since you have

$$S_p=\frac{\prod_{n=1}^p (2n) }{\prod_{n=1}^p (2n-1) }=\frac{2^p \Gamma (p+1) } {\frac{2^p \Gamma \left(p+\frac{1}{2}\right)}{\sqrt{\pi }} }=\sqrt{\pi }\,\,\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}$$ Take logarithms, use Stirling approximation and continue with Taylor series for large $p$ $$A=\log (\Gamma (p+1))-\log \left(\Gamma \left(p+\frac{1}{2}\right)\right)$$ $$A=\frac{1}{2}\log (p)+\frac{1}{8 p}-\frac{1}{192 p^3}+O\left(\frac{1}{p^5}\right)$$ $$\frac{ \Gamma (p+1)}{\Gamma \left(p+\frac{1}{2}\right)}=e^A=\sqrt{p}\left(1+\frac{1}{8 p}+\frac{1}{128 p^2}+O\left(\frac{1}{p^3}\right) \right)$$

$$S_p >\sqrt{p\pi}\left(1+\frac{1}{8 p}\right) $$ Using it for $p=5000$ gives $125.334547017$ while the exact value is $125.334547056$

Claude Leibovici
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