If n is a positive integer, prove that $\dfrac{1}{2\sqrt{n+1}} < \dfrac{1\cdot3\cdot5\ldots(2n-1)}{2\cdot4\cdot6\ldots2n}$
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A stronger inequality as $$\frac{1}{2\sqrt{n}} \ge \frac{1}{2\sqrt{n+1}}$$ the result follows – Albus Dumbledore Mar 06 '21 at 03:20
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Have you tried anything? What are you studying, that is, what tools might you be expected to use? – robjohn Mar 06 '21 at 03:20
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I was trying to use A.M > G.M and then put the prod operation and I thought I could get some way in middle, where I can reverse the sign of inequality. But that gave me, $\dfrac{1\cdot3\cdot5\ldots(2n-1)}{2\cdot4\cdot6\ldots2n}< \dfrac{1}{\sqrt{2n+1}}$ – Subham Karmakar Mar 06 '21 at 03:33
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1Thanks @Albus Dumbledore, I understood how to prove that. – Subham Karmakar Mar 06 '21 at 03:37