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What would be the right way to solve this by induction proof?

$$\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$$

This is what I've done (reference https://www.slader.com/discussion/question/prove-that-12n-1-3-5-2n-12-4-2n-whenever-n-is-a-positive-integer/#):

  1. Show that $S\left(n+1\right)$ by induction proof. This is $$\frac{1}{2(n+1)}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n+1)}{2+4+6+\ldots+2(n+2)}$$

Multiplying both sides of the equation $\frac{2n\text{·}(2(n+1)-1)}{2n\text{·}(2(n+1)-1)}=\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$

$$\frac{1}{2n}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}\times\frac{2n\text{·}(2(n+1)-1}{2n\text{·}(2(n+1)-1}$$

$$\frac{1}{2n}\times\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$$ Rewriting we have the following \begin{array}{c} \frac{2n+1}{2n(2n+1)}\\ \frac{1}{2n+1}+\frac{1}{2n(2n+1)} \end{array}

$$\frac{1}{2n}\leq\frac{1}{2n+1}+\frac{1}{2n(2n+1)}$$

Diego Gallo
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  • If possible, could you translate to English? This will allow most people to be able to read the question. – Simply Beautiful Art Oct 29 '20 at 23:25
  • ¡Bienvenido a Math StackExchange! ¿Puede proporcionar una traducción al inglés de esta publicación? (Welcome to the Math StackExchange! Can you please provide an English translation of this post?) – N. Bar Oct 29 '20 at 23:27
  • Are you sure those are supposed to be addition signs in the denominator as opposed to multiplication? – JimmyK4542 Oct 30 '20 at 00:00

2 Answers2

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$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{2+4+6+\ldots+2n}\ge \frac{1}{2n}\tag{1} $$ can be simplified as $$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{n(n+1)}\ge \frac{1}{2n} $$ and then $$1\cdot3\cdot5\cdot\ldots\cdot(2n-1)\ge\frac{n+1}{2}\tag{2}$$ for $n=1$ we have $1\ge 1$ true.

Now suppose $(2)$ is true and let us prove it for $(n+1)$.

$$[1\cdot3\cdot5\cdot\ldots\cdot(2n-1)](2n+1)\ge \frac{n+1}{2}\cdot(2n+1)\ge\frac{n+1+1}{2}=n+1$$

Raffaele
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0

Let $\varphi, \psi:\mathbb N\to\mathbb N$ given by \begin{aligned} \varphi(n) &= \sum_{k=1}^n 2k = n(n+1)\\ \psi(n) &= \prod_{k=1}^n (2k-1)\\ \end{aligned} and notice that \begin{aligned} \varphi(n+1) &= \frac{n+2}n\varphi(n)\\ \psi(n+1) &= (2n+1)\psi(n)\\ \end{aligned}

If you assume that, for a certain $n\in\mathbb N$, the following is true $$\frac 1{2n}\le \frac{1\times 3\times\ldots\times (2n-1)}{2 + 4 + \ldots + 2n} = \frac{\psi(n)}{\varphi(n)},$$ then you may write \begin{aligned} \frac 1{2(n+1)} & = \frac{2n}{2(n+1)}\frac1{2n}\\ & \le \frac{2n}{2(n+1)}\frac{\psi(n)}{\varphi(n)}\\ & = \frac{2n}{2(n+1)}\frac1{2n+1}\frac{n+2}{n}\frac{\psi(n+1)}{\varphi(n+1)}\\ & = \frac{n+2}{(n+1)(2n+1)}\frac{\psi(n+1)}{\varphi(n+1)}\\ & \le \frac{\psi(n+1)}{\varphi(n+1)}, \end{aligned}

where the last inequality follows from the fact that

$$ \frac{n+2}{(n+1)(2n+1)} = \frac{1}{2n+1}\left(1 + \frac1{n+1}\right) \le \frac 12\left(1 + \frac 12\right) = 1. $$

foreignvol
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