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I am lookinf for an idea to prove this inequality: $$\frac{1}{2}\cdot \frac{3}{4}\cdot\frac{5}{6}...\frac{2n-1}{2n}<\frac{1}{\sqrt{2n+1}}$$ without induction.

I can use induction to prove it .Thanks in advance for your hint,solution and idea.

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$$\underbrace{\frac12\cdot\frac23}_{<\left(\frac12\right)^2}\cdot\underbrace{\frac34\cdot\frac45}_{<\left(\frac34\right)^2}\cdots\underbrace{\frac{2n-1}{2n}\cdot\frac{2n}{2n+1}}_{<\left(\frac{2n}{2n-1}\right)^2}=\frac1{2n+1}$$

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  • $\begingroup$ I think it's exactly my and the Lord Shark the Unknown's solution! Because you used $\frac{2k}{2k+1}>\frac{2k-1}{2k}$, which also happens in our solutions. $\endgroup$ – Michael Rozenberg Aug 12 '17 at 13:40
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$$\frac{1}{2}\cdot \frac{3}{4}\cdot\frac{5}{6}...\frac{2n-1}{2n}=\sqrt{\frac{3}{4}\cdot\frac{3\cdot5\cdot5\cdot7...\cdot(2n-3)(2n-1)\cdot(2n-1)}{4^2\cdot6^2...(2n-2)^2\cdot4n^2}}<$$ $$<\sqrt{\frac{3(2n-1)}{16n^2}}<\frac{1}{\sqrt{2n+1}}$$

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  • $\begingroup$ :Can you explain the last step ? $\endgroup$ – Khosrotash Aug 12 '17 at 13:34
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    $\begingroup$ @Khosrotash It's $3(4n^2-1)<16n^2$. $\endgroup$ – Michael Rozenberg Aug 12 '17 at 13:38
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This is the same as $$\frac{1\times 3\times 3\times 5\cdots\times(2n-1)(2n+1)}{2\times 2\times 4\times 4\times\cdots\times 2n\times 2n}<1$$ or $$\prod_{j=1}^n\frac{(2j-1)(2j+1)}{(2j)^2} =\prod_{j=1}^n\frac{4j^2-1}{4j^2}<1.$$

Does this count as induction?

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Please do not take this answer too seriously: it clearly is overkill, meant to show a (hopefully) interesting technique. We have: $$f(x)=\sum_{n\geq 0}\frac{x^{2n}}{16^n}\binom{2n}{n}^2 = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-x^2\sin^2 t}}=\frac{2}{\pi}\int_{0}^{1}\frac{du}{\sqrt{1-u^2}\sqrt{1-x^2 u^2}} $$ $$ g(x)=\sum_{n\geq 0}\frac{x^{2n}}{2n+1} = \frac{\text{arctanh} x}{x} $$ i.e. two analytic functions in $(-1,1)$, whose Taylor series at the origin only has non-negative coefficients. We would like to show that for any $m\in\mathbb{N}^+$ the inequality $[x^{2m}]\,f(x)<[x^{2m}]\,g(x)$ holds, or $f\ll g$ for brevity. For such a purpose, we may apply the operator $\delta:h(x)\mapsto (1-x^2)\cdot\frac{d}{dx}\left(x\cdot h(x)\right)$ to both $f$ and $g$. We have $\delta g(x)=1$ and $$ \delta f(x) = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{1-x^2}{\left(1-x^2\sin^2 t\right)^{3/2}}\,dt $$ is an analytic function in $(-1,1)$ whose Taylor series at the origin has the form $1-\frac{x^2}{4}-\frac{3x^4}{64}-\ldots$ (by integration by parts, only non-positive coefficients follow).
$\delta^{-1}$ acts in the following way:

$$ \delta^{-1}: H(x)\mapsto \frac{1}{x}\int_{0}^{x} H(u)\cdot\left(1+u^2+u^4+\ldots\right)\,du $$ hence we have $f\ll g$ as wanted. Essentially, a termwise-comparison of Taylor coefficients can be performed by noticing that $f$ and $g$ are strongly convex functions fulfilling differential equations with a similar structure.

Another approach by (log-)convexity is just to exploit Gautschi's inequality.

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