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Prove $\frac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\frac{1}{\sqrt{2n+1}}$, $n\ge 1$.

I begin by letting $n=1$ then $\frac{1}{2}<\frac{1}{\sqrt{3}}$. Then assume $\frac{1\cdot 3\cdots(2k-1)}{2\cdot 4 \cdots(2k)}<\frac{1}{\sqrt{2k+1}}$, for some $k\ge 1$. Then if I multiply both sides by $\frac{(2(k+1)-1)}{2(k+1)}$ I get $\frac{1\cdot 3\cdots(2(k+1)-1)}{2\cdot 4 \cdots2(k+1)}<\frac{1}{\sqrt{2k+1}}\cdot \frac{(2(k+1)-1)}{2(k+1)}$. So I have what I need on the left hand side, but I'm not sure how to continue on the right hand side. I realized that if I show that $\frac{(2(k+1)-1)}{2(k+1)}<\frac{\sqrt{2k+1}}{\sqrt{2(k+1)+1}}$ then I would be done. This however leads to the same dilemma in that I can easily make the left the way I need, but not the right. So there must be a trick that I am missing.

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    $\begingroup$ Have you tried cross-multiplying, squaring, and simplifying to check whether or not $\frac{2(k+1)-1}{2(k+1)} < \frac{\sqrt{2k+1}}{\sqrt{2(k+1)+1}}$? This would be the most direct way to continue your argument. $\endgroup$ – Dan Jan 26 '16 at 16:45
  • $\begingroup$ What @Dan said. You might want to put $j=k+1$ to simplify the algebra. $\endgroup$ – TonyK Jan 26 '16 at 16:47
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    $\begingroup$ I will work on that, I was looking for a much more implicit trick and failed to see the obvious one $\endgroup$ – Burgundy Jan 26 '16 at 16:47
  • $\begingroup$ @Burgundy: It's straightforward to show that $\frac{2x-1}{2x} < \frac{\sqrt{2x-1}}{\sqrt{2x+1}}$ (which is what you have) directly for $x = k+1$ and $k$ a natural number. You could also try something slicker, e.g., noting that $2\cdot 4 \cdots (2n) = 2^n n!$ or expanding the fractions into products like in the answer below, but I was attempting to help you finish your argument. All are valid options. :) $\endgroup$ – Dan Jan 26 '16 at 16:52
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Here is the inductive step: from $\;\dfrac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\dfrac{1}{\sqrt{2n+1}}$, you deduce $$\dfrac{1\cdot 3\cdots(2n-1)(2n+1)}{2\cdot 4 \cdots(2n)(2n+2)}<\dfrac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2},$$ hence it is enough to prove $\;\dfrac{1}{\sqrt{2n+1}}\dfrac{2n+1}{2n+2}<\dfrac1{\sqrt{2n+3}}$. This is equivalent to $$\sqrt{(2n+1)(2n+3)}<2n+2,$$ which is the A.G.M. inequality (case of strict inequality).

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First observe that $$\frac{2n-1}{2n}<\frac{2n}{2n+1}\qquad\text{since }(2n-1)(2n+1)=4n^2-1<4n^2=(2n)(2n)$$ Then \begin{align} \left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)^2&<\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)\left(\prod_{k=1}^{n}\frac{2k}{2k+1}\right)\\ &=\prod_{k=1}^{n}\frac{2k-1}{2k+1}\qquad\text{this product is "telescopic"}\\ &=\frac{1}{2n+1} \end{align} So $$\prod_{k=1}^{n}\frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$$

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